GIFT   OF 
MICHAEL  REE&E 


MECHANICS  OF  ENGINEERING. 

12] 
[FLUIDS.] 

A  TREATISE  ON  HYDRAULICS  AND  PNEUMATICS. 
FOR    USE  IN  TECHNICAL  SCHOOLS. 


BY 

IRVING  P.   CHURCH,  C.E., 

ASSISTANT  PROFESSOR  OF  CIVIL  ENGINEERING,  CORNELL  UNIVERSITY. 
(IN  CHARGE  OF  APPLIED  MECHANICS.) 


NEW   YORK: 

JOHN    WILEY    &    SONS, 
15  ASTOR   PLACE. 

1889. 


, 


Copyright,  1889, 

BY 

IRVING  P.  CHURCH. 


PnuMMOND  &  NBU,  FERRIS  BROS., 

Klectrotypera,  Printers, 

to  7  Hague  Street,  828  Pearl  Street, 
New  York.  New  York. 


G.  f. 


PREFACE. 


THE  same  general  design  has  been  kept  in  view  in  the  prep- 
aration of  the  following  work  as  in  the  preceding  pages  on 
Solids,  viz.  :  to  combine  clearness  and  consistency  in  the  setting 
forth  and  illustration  of  theoretical  principles ;  to  provide  nu- 
merous and  fully-lettered  diagrams,  in  which  in  most  cases  the 
notation  of  the  accompanying  text  can  be  apprehended  at  a 
glance  ;  and  to  invite  close  attention  to  the  proper  use  of  systems 
of  units  in  numerical  examples,  the  latter  being  introduced  very 
copiously  and  with  detailed  explanations. 

Advantage  has  been  taken  of  the  results  of  the  most  reeent 
experimental  investigations  in  Hydraulics  in  assigning  values  of 
the  numerous  coefficients  necessary  in  this  science.  The  re- 
searches of  Messrs.  Fteley  and  Stearns  in  1880  and  of  M.  Bazin 
in  1887  on  the  flow  of  water  over  weirs,  and  of  Mr.  Clemens 
Herschel  in  testing  his  invention  the  "Venturi  Water-meter/' 
are  instances  in  point ;  as  also  some  late  experiments  on  the 
transmission  of  compressed  air  and  of  natural  gas. 

Though  space  has  forbidden  dealing  at  any  great  length  with 
the  action  of  fluid  motors,  'sufficient  matter  is  given  in  treating 
of  the  mode  of  working  of  steam,  gas,  and  hot-air  engines,  air- 
compressors,  and  pumping-engines,  together  with  numerical  ex- 
amples, to  be  of  considerable  advantage,  it  is  thought,  to  students 
not  making  a  specialty  of  mechanical  engineering. 

Special  acknowledgment  is  due  to  Col.  J.  T.  Fanning,  the 
well-known  author  of  "  Hydraulic  and  Water-supply  Engineer- 
ing," for  his  consent  to  the  use  of  an  abridgment  of  the  table  of 
coefficients  for  friction  of  water  in  pipes,  given  in  that  work ; 
and  to  Prof.  C.  L.  Crandall,  of  this  university,  for  permission  to 
incorporate  the  chapter  on  Retaining-walls. 

References  to  original  research  in  the  Hydraulic  Laboratory  of 
the  Civil  Engineering  Department  at  this  institution  will  be 
found  on  pp.  694  and  729. 

CORNELL  UNIVERSITY,  ITHACA,  N.  Y.,  May  1889. 

i 

Hi 


CONTENTS. 


PAKT  IV.— HYDKAULICS. 

CHAPTER  I.— DEFINITIONS.     FLUID  PRESSURE.    HYDRO- 
STATICS BEGUN. 

PAGE 

§§  406-417.  Perfect  fluids.  Liquids  and  Gases.  Principle  of  ' '  Equal 

Transmission  of  Pressure."  Non-planar  Pistons. . .  515 

§§  418-427.  Hydraulic  Press.  Free  Surface  of  Liquid.  Barometers 
and  Manometers.  Safety-valves.  Strength  of  Thin 
Hollow  Cylinders  against  Bursting  and  Collapse. . .  526 

CHAPTER  II.— HYDROSTATICS    CONTINUED.     PRESSURE   OF 
LIQUIDS  IN  TANKS  AND  RESERVOIRS. 

§§428-434.  Liquid  in  Motion,  but  in  "Relative  Equilibrium." 
Pressure  on  Bottom  and  Sides  of  Vessels.  Centre 
of  Pressure  of  Rectangles,  Triangles,  etc 540 

§§  435-444.  Stability  of  Rectangular  and  Trapezoidal  Walls  against 
Water  Pressure.  High  Masonry  Dams.  Proposed 
Quaker  Bridge  Dam.  Earthwork  Dam.  Water 
Pressure  on  both  Sides  of  a  Gate 554 

CHAPTER  III.-EARTH  PRESSURE  AND  RETAINING  WALLS. 

§§  445-455.  Angle  of  Repose.  Wedge  of  Maximum  Thrust.  Geo- 
metrical Constructions.  Resistance  of  Retaining 
Walls.  Results  of  Experience 572 

CHAPTER  IV.— HYDROSTATICS  CONTINUED.     IMMERSION 
AND  FLOTATION. 

§§456-460.  Buoyant  Effort.  Examples  of  Immersion.  Specific 

Gravity.  Equilibrium  of  Flotation.  Hydrometer..  586 

§§  461-465.  Depth  of  Flotation.  Draught  and  Angular  Stability  of 

Ships.  The  Metacentre 592 


VI 


CONTENTS. 


CHAPTER  V.— HYDROSTATICS  CONTINUED. 
FLUIDS. 


GASEOUS 


§§466-478.  Thermometers.  Absolute  Temperature.  Gases  and  Va- 
pors. Critical  Temperature.  Law  of  Charles. 
Closed  Air-manometer.  Mariotte's  Law.  Mixture  of 
Gases.  Barometric  Levelling.  Adiabatic  Change..  604 

§§479-489.  Work  Done  in  Steam-engine  Cylinders.  Expanding 
Steam.  -Graphic  Representation  of  Change  of  State 
of  Gas.  Compressed-air  Engine.  Air-compressor. 
Hot-air  Engines.  Gas-engines.  Heat- efficiency. 
Duty  of  Pumpiug-engines.'  Buoyant  Effort  of  the 
Atmosphere 624 

CHAPTER  VI.— HYDRODYNAMICS  BEGUN.     STEADY  FLOW 
OF  LIQUIDS  THROUGH  PIPES  AND  ORIFICES. 


§§  489a-495. 

\ 
§§  496-500. 

§§  501-508. 
§§  509-513. 

§§  513o-518. 
§§  519-526. 

S   527-536. 


Phenomena  of  a  "  Steady  Flow."  Bernoulli's  Theorem 
for  Steady  Flow  without  Friction,  and  Applications. 
Orifice  in  "  Thin  Plate" 646 

Rounded  Orifice.  Various  Problems  involving  Flow 
through  Orifices.  Jet  from  Force-pump.  Velocity 
and  Density;  Relation.  Efflux  under  Water.  Efflux 
from  Vessel  in  Motion.  Barker's  Mill 663 

Efflux  from  Rectangular  and  Triangular  Orifices.  Pon- 
celet's  Experiments.  Perfect  and  Complete  Con- 
traction, etc.  Overfall  Weirs.  Experiments  of 
Francis,  Fteley  and  Stearns,  and  Bazin.  Short 
Pipes  or  Tubes : . .  672 

Conical  Tubes.  Venturi's  Tube.  Fluid  Friction. 
Froude's  Experiments.  Bernoulli's  Theorem  with 
Friction.  Hydraulic  Radius.  Loss  of  Head.  Prob- 
lems involving  Friction  Heads  in  Pipes.  Accumu- 
lator   692 

Loss  of  Head  in  Orifices  and  Short  Pipes.  Coefficient 
of  Friction  of  Water  in  Pipes.  Fanning 's  Table. 
Petroleum  Pumping.  Flow  through  Long  Pipes. .  70& 

Chezy's  Formula.  Fire-engine  Hose.  Pressure-energy. 
Losses  of  Head  due  to  Sudden  Enlargement  of  Sec- 
tion; Borda's  Formula.  Diaphragm  in  Pipe.  Ven- 
turi  Water-meter 714 

Sudden  Diminution  of  Section.  Losses  of  Head  due  to 
Elbows,  Bends,  Valve-gates,  and  Throttle-valves. 
Examples,  Prof.  Bellinger's  Experiments  on  Elbows. 
Siphons.  Branching  Pipes.  Time  of  Emptying 
Vessels  of  Various  Forms;  Prisms,  Wedges.  Pyra- 
mids, Cones,  Paraboloids,  Spheres,  Obelisks,  and 
Volumes  of  Irregular  Form  using  Simpson's  Rule. .  727 


CONTENTS.  Vil 

PAGE 

CHAPTER  VII.— HYDRODYNAMICS,  CONTINUED;  STEADY 
FLOW  OF  WATER  IN  OPEN  CHANNELS. 

:§§  538-542a.  Nomenclature.  Velocity  Measurements  and  Instru- 
ments for  the  same.  Ritchie-Haskell  Direction 
Current-meter.  Change  of  Velocity  with  Depth. 
Pitot's  Tube.  Hydrometric  Pendulum.  Wolt- 
mann's  Mill.  Gauging  Streams.  Chezy's  Formula 
for  Uniform  Motion  in  Open  Channel.  Experi- 
ments   749 

.§§  5425-547.  Kutter's  Formula.  Sections  of  Least  Resistance.  Trape- 
zoidal Section  of  Given'  Side  Slope  and  Minimum 
Friction.  Variable  Motion  in  Open  Channel. 
Bends.  Formula  introducing  Depths  at  End  Sec- 
tions. Backwater 759 

CHAPTER  VIII.— DYNAMICS  OF  GASEOUS  FLUIDS. 

§§  548-556.  Theorem  for  Steady  Flow  of  Gases  without  Friction. 
Flow  through  Orifices  by  Water-formula;  with 
Isothermal  Expansion;  with  Adiabatic  Expansion. 
Maximum  Flow  of  Weight.  Experimental  Co- 
efficients for  Orifices  and  Short  Pipes.  Flow  con- 
sidering Velocity  of  Approach 773 

§§  557-561&.  Transmission  of  Compressed  Air  through  Long  Pipes. 
Experiments  in  St.  Gothard  Tunnel.  Pipes  of  Vari- 
able Diameter.  The  Piping  of  Natural  Gas 786 

CHAPTER  IX.— IMPULSE  AND  RESISTANCE  OF  FLUIDS. 

§§  563-569.  Reaction  of  a  Jet  of  Liquid.  Impulse  of  Jet  on  Curved 
Vanes,  Fixed  and  in  Motion.  Pitot's  Tube.  The 
California  "Hurdy-gurdy."  Impulse  on  Plates. 
Plates  Moving  in  a  Fluid.  Plates  in  Currents  of 
Fluid 798 

•§§  570-575.  Wind-pressure.  Smithsonian  Scale.  Mechanics  of  the 
Sail-boat.  Resistance  of  Still  Water  to  Immersed 
Solids  in  Motion.  Spinning  Ball,  Deviation  from 
Vertical  Plane.  Robinson's  Cup  anemometer.  Re- 
sistance of  Ships.  Transporting  Power  of  a  Cur- 
rent.., .  818 


PART  IV. 

HYDRAULICS. 


CHAPTEK  I. 

DEFINITIONS— FLUID  PRESSURE— HYDROSTATICS  BEGUN. 

406.  A  Perfect  Fluid  is  a  substance  the  particles  of  which 
are  capable  of  moving  upon  each  other  with  the  greatest  free 
dom,  absolutely  without  friction,  and  are  destitute  of  mutual 
attraction.     In  other  words,  the  stress  between  any  two  con- 
tiguous portions  of  a  perfect  fluid  is  always  one  of  compression 
and  normal  to  the  dividing  surface  at  every  point ;  i.e.,  no 
shear  or  tangential  action  can  exist  on  any  imaginary  cutting 
plane. 

Hence  if  a  perfect  fluid  is  contained  in  a  vessel  of  rigid  ma- 
terial the  pressure  experienced  by  the  walls  of  the  vessel  is 
normal  to  the  surface  of  contact  at  all  points. 

For  the  practical  purposes  of  Engineering,  water,  alcohol, 
mercury,  air,  steam,  and  all  gases  may  be  treated  as  perfect 
fluids  within  certain  limits  of  temperature. 

407.  Liquids  and  Gases. — A  fluid  a  definite  mass  of  which 
occupies  a  definite  volume  at  a  given  temperature,  and  is  in- 
capable both  of  expanding  into  a  larger  volume  and  of  being 
compressed  into  a  smaller  -volume  at  that  temperature,  is  called 
a  Liquid,  of  which  water,  mercury,  etc.,  are  common  examples ; 
whereas  a  Gas  is  a  fluid  a  mass  of  which  is  capable  of  almost 
indefinite  expansion  or  compression,  according  as  the  space 
within  the  confining  vessel  is  made  larger  or  smaller,  and  al- 
ways tends  to  fill  the  vessel,  which  must  therefore  be  closed  in 
every  direction  to  prevent  its  escape. 

515 


516  MECHANICS    OF   ENGINEERING. 

Liquids  are  sometimes  called  inelastic  fluids,  and  gases 
elastic  fluids. 

408.  Remarks. — Though  practically  we  may  treat  all  liquids 
as  incompressible,  experiment  shows  them  to  be  compressible 
to  a  slight  extent.  Thus,  a  cubic  inch  of  water  under  a  pres- 
sure of  15  Ibs.  on  each  of  its  six  faces  loses  only  fifty  millionths 
(0.000050)  of  its  original  volume,  while  remaining  at  the  same 
temperature;  if  the  temperature  be  sufficiently  raised,  how- 
ever, its  bulk  will  remain  unchanged  (provided  the  initial  tem- 
perature is  over  40°  Fahr.).  Conversely,  by  heating  a  liquid  in 
a  rigid  vessel  completely  filled  by  it,  a  great  bursting  pressure 
may  be  produced.  The  slight  cohesion  existing  between  the 
particles  of  most  liquids  is  too  insignificant  to  be  considered  in 
the  present  connection. 

The  property  of  indefinite  expansion,  on  the  part  of  gases, 
by  which  a  confined  mass  of  gas  can  continue  to  fill  a  confined 
space  which  is  progressively  enlarging,  and  exert  pressure 
against  its  walls,  is  satisfactorily  explained  by  the  "  Kinetic 
Theory  of  Gases,"  according  to  which  the  gaseous  particles  are 
perfectly  elastic  and  in  continual  motion,  impinging  against 
each  other  and  the  confining  walls.  Nevertheless,  for  prac- 
tical purposes,  we  may  consider  a  gas  as  a  continuous  sub- 
stance. 

Although  by  the  abstraction  of  heat,  or  the  application  of 
great  pressure,  or  both,  all  known  gases  may  be  reduced  to 
liquids  (some  being  even  solidified);  and  although  by  con- 
verse processes  (imparting  heat  and  diminishing  the  pressure) 
liquids  may  be  transformed  into  gases,  the  range  of  tempera- 
ture and  pressure  in  all  problems  to  be  considered  in  this  work 
is  supposed  kept  within  such  limits  that  no  extreme  changes  of 
state,  of  this  character,  take  place.  A  gas  approaching  the 
point  of  liquefaction  is  called  a  Vapor. 

Between  the  solid  and  the  liquid  state  we  find  all  grades  of 
intermediate  conditions  of  matter.  For  example,  some  sub- 
stances are  described  as  soft  and  plastic  solids,  as  soft  putty, 
moist  earth,  pitch,  fresh  mortar,  etc.;  and  others  as  viscous  and 
sluggish  liquids,  as  molasses  and  glycerine.  In  sufficient  bulk, 


DEFINITIONS— FLUID   PRESSURE— HYDROSTATICS.    517 

however,  the  latter  may  still  be  considered  as  perfect  fluids. 
Even  water  is  slightly  viscous. 

409,  Heaviness  of  Fluids. — The  weight  of  a  cubic  unit  of  a 
homogeneous  fluid  will  be  called  its  heaviness,  or  rate  of 
weight  (see  §  7),  and  is  a  measure  of  its  density.  Denoting  it 
by  y,  and  the  volume  of  a  definite  portion  of  the  fluid  by  V, 
we  have,  for  the  weight  of  that  portion, 


G=  Vy. 


(1) 


This,  like  the  great  majority  of  equations  used  or  derived  in 
this  work,  is  of  homogeneous  form  (§  6),  i.e.,  admits  of  any  sys- 
tem of  units.  E.g.,  in  the  metre-kilogram-second  system,  if  y 
is  given  in  kilos,  per  cubic  metre,  V  must  be  expressed  in 
cubic  metres,  and  G  will  be  obtained  in  kilos.;  and  similarly 
in  any  other  system.  The  quality  of  y,  =  G  -f-  F,  is  evidently 
one  dimension  of  force  divided  by  three  dimensions  of  length. 

In  the  following  table,  in  the  case  of  gases,  the  temperature 
and  pressure  are  mentioned  at  which  they  have  the  given 
heaviness,  since  under  other  conditions  the  heaviness  would  be 
different ;  in  the  case  of  liquids,  however,  for  ordinary  pur- 
poses the  effect  of  a  change  of  temperature  may  be  neglected 
(within  certain  limits). 


HEAVINESS  OF  VARIOUS  FLUIDS.* 
[In  ft.  Ib.  sec.  system;  y  =  weight  in  Ibs.  of  a  cubic  foot.] 


Liquids. 


J  At  temp,  of  melting  ice;  and  14.7 
1     Ibs.  per  sq.  in.  tension. 


Freshwater,  y=  62.5 

Sea  water 64.0 

Mercury 848. 7 

Alcohol 49.3 

Crude  Petroleum,  about 55.0 

(N.B. — A  cubic  inch  of  water 
weighs  0.036034  Ibs.;  and  a  cubic 
foot  1000  av.  oz.) 


Atmospheric  Air 0. 08076 

Oxygen 0.0892 

Nitrogen 0.07'86 

Hydrogen 0.0056 

Illuminating  )  from 0.0300 

Gas,  fto 0.0400 

Natural  Gas,  about 0.0500 


*  See  Trautwine's  Civ.  Engineer's  Pocket  Book  for  an  extended  table — 
p.  380,  edition  of  1885. 


518 


MECHANICS    OF   ENGINEERING. 


For  use  in  problems  where  needed,  values  for  the  heaviness 
of  pure  fresh  water  are  given  in  the  following  table  (from 
Rossetti)  for  temperatures  ranging  from  freezing  to  boiling ; 
as  also  the  relative  density,  that  at  the  temperature  of  maxi- 
mum density,  39°. 3  Fahr.  being  taken  as  unity.  The  temper- 
atures are  Fahr.,  and  y  is  in  Ibs.  per  cubic  foot. 


Temp. 

Rel. 
Dens. 

Y. 

Temp. 

Rel. 
Dens. 

y. 

Temp. 

Rel. 
Dens. 

V- 

32° 

.99987 

62.416 

60° 

.99907 

62.366 

140° 

.98338 

61.386 

35° 

.99996 

62.421 

70° 

.99802 

62.300 

150° 

.98043 

61.203 

39°.  3 

1.00000 

62.424 

80° 

.99669 

62.217 

160° 

.97729 

61.006 

40° 

.99999 

62.423J 

90° 

.99510 

62.118 

170° 

.97397 

60.799 

43° 

.99997 

62.422 

100° 

.99318 

61.998 

180° 

.97056 

60.586 

45° 

.99992 

62.419 

110° 

.99105 

61.865 

190° 

.96701 

60.365 

50° 

.99975 

62.408 

120° 

.98870 

61.719 

200° 

.96333 

60.135 

55° 

.99946 

62.390 

130° 

.98608 

61.555 

212° 

.95865 

59.843 

EXAMPLE  1.  What  is  the  heaviness  of  a  gas,  432  cub.  in.  of 
which  weigh  0.368  ounces?     Use  ft.-lb.-sec.  system. 

432  cub.  in.  =  \  cub.  ft.  and  0.368  oz.  =  0.023  Ibs. 


=  0.092  Ibs.  per  cub.  foot. 


EXAMPLE  2.  Required  the  weight  of  a  right  prism  of  mer- 
cury of  1  sq.  inch  section  and  30  inches  altitude. 

OA 

V  =30  X  1  =  30  cub.  in.  =  -^--  cub.  feet  ;  while  from  the 

1728 

table,  y  for  mercury  =  848.7  Ibs.  per  cub.  ft. 

OA 

.-.  its  weight  =  G  =  Vy  =  -%-  X  848.7  =  14.73  Ibs. 


410.  Definitions.  —  By  Hydraulics  we  understand  the  me- 
chanics of  fluids  as  utilized  in  Engineering.  It  may  be  divided 
into 

Hydrostatics,  treating  of  fluids  at  rest  ;  and 

Hydrodynamics  (or  Hydrokinetics),  which  deals  with  fluids 
in  motion.  (The  name  Pneumatics  is  sometimes  used  to  cover 
both  the  statics  and  dynamics  of  gaseous  fluids.) 


DEFINITIONS—  FLUID   PRESSURE—  HYDROSTATICS.    519 

[Rankine's  nomenclature  has  been  adopted  in  the  present 
work.  Some  recent  writers  use  the  term  Hydromechanics  for 
mechanics  of  fluids,  subdividing  it  into  Hydrostatics  and 
Hydrokinetics,  as  above  ;  they  also  use  the  term  Dynamics  to 
embrace  both  of  the  two  divisions  called  Statics  and  Dynamics 
by  Rankine,  which  by  them  are  called  Statics  and  Kinetics  re- 
spectively. Though  unusual,  perhaps,  the  term  Hydraulics  is 
here  used  to  cover  the  applied  Mechanics  of  Fluids  as  well  as 
of  Liquids.] 

Before  treating  separately  of  liquids  and  gases,  a  few  para- 
graphs will  be  presented  applicable  to  both  kinds  of  fluids. 

411.  Pressure  per  Unit  Area,  or  Intensity  of  Pressure.  —  As  in 

§  180  in  dealing  with  solids,  so  here  with  fluids  we  indicate  the 
pressure  per  unit  area  between  two  contiguous  portions  of 
fluid,  or  between  a  fluid  and  the  wall  of  the  containing  vessel, 
by  p,  so  that  if  dP  is  the  total  pressure  on  a  small  area  dF, 
we  have 

>-£  ........  «• 

as  the  pressure  per  unit  area,  or  intensity  of  pressure  (often, 
though  ambiguously,  called  the  tension  in  speaking  of  a  gas) 
on  the  small  surface  dF.  If  pressure  of  the  same  intensity 
exists  over  a  finite  plane  surface  of  area  =  F,  the  total  pres- 
sure on  that  surface  is 

P  =  fpdF=pfdF=  Fp,  ] 

P  [     •    •    •    .    (2)" 

=    >. 


(!N".B.  —  For  brevity  the  single  word  "  pressure"  will  some- 
times be  used,  instead  of  intensity  of  pressure,  where  no  am- 
biguity can  arise.)  Thus,  it  is  found  that,  under  ordinary  con- 
ditions at  the  sea  level,  the  atmosphere  exerts  a  normal  pressure 
(normal,  because  fluid  pressure)  on  all  surfaces,  of  an  intensity 
of  about  p  —  14.7  Ibs.  per  sq.  inch  (=  2116.  Ibs.  per  sq.  ft.). 
This  intensity  of  pressure  is  called  one  atmosphere.  For  ex- 


520 


MECHANICS    OF   ENGINEERING. 


ample,  the  total  atmospheric  pressure  on  a  surface  of  100  sq. 
in.  is  [inch,  lb.,  sec.] 

P  =  Fp  =  100  X  14.7  =  1470  Ibs.     (=  0.735  tons.) 

The  quality  of  p  is  evidently  one  dimension  of  force  divid- 
ed by  two  dimensions  of  length. 

412.  Hydrostatic  Pressure;  per  Unit  Area,  in  the  Interior  of  a 
Fluid  at  Rest. — In  a  body  of  fluid  of  uniform  heaviness,  at 
rest,  it  is  required  to  find  the  mutual  pressure  per  unit  area  be- 
tween the  portions  of  fluid  on  opposite  sides  of  any  imaginary 
cutting  plane.  As  customary,  we  shall  consider  portions  of 
the  fluid  as  free  bodies,  by  supplying  the  forces  exerted  on 
them  by  all  contiguous  portions  (of  fluid  or  vessel  wall),  also- 
those  of  the  earth  (their  weights),  and  then  apply  the  condi- 
tions of  equilibrium. 

First)  cutting  plane  horizontal. — Fig.  451  shows  a  body  of 
homogeneous  fluid  confined  in  a  rigid 
vessel  closed  at  the  top  with  a  small  air- 
tight but  frictionless  piston  (a  horizontal 
disk)  of  weight  =  G  and  exposed  to  at- 
mospheric pressure  (— ^?a  per  unit  area) 
on  its  upper  face.  Let  the  area  of  piston- 
face  be  =  F.  Then  for  the  equilibrium 
of  the  piston  the  total  pressure  between 
its  under  surface  and  the  fluid  at  0  must 
be 


FIG.  451. 


and  hence  the  intensity  of  this  pressure  is 


(1) 


It  is  now  required  to  find  the  intensity,^,  of  fluid  pressure 
between  the  portions  of  fluid  contiguous  to  the  horizontal  cut- 
ting plane  BC&i  a  vertical  distance  =  h  vertically  below  the  pis- 
ton O.  In  Fig.  452  we  have  as  a  free  body  the  right  parallelo- 


FLUID   PRESSURE. 


olliil 


piped  OBC  oi  Fig.  451  with  vertical  sides  (two  ||  to  paper  and 
four  ~]  to  it).  The  pressures  acting  on  its  six  faces  are  normal 
to  them  respectively,  and  the  weight  of  the  prism  is  —  vol. 
Xy  =  Fhy,  supposing  y  to  have  the  same  value  at  all  parts  of 
the  column  (which  is  practically  true  for  any  height  of  liquid 
and  for  a  small  height  of  gas).  Since  the  ^-^ — ^n 

prism  is  in  equilibrium  under  the  forces 
shown  in  the  figure,  and  would  still  be  so 
were  it  to  become  rigid,  we  may  put  (§  36) 
.2  (vert,  compons.)  =  0  and  hence  obtain 

Fp-Fp.-  Fhy  =  0.  .    .    (2)J 

(In  the   figure  the  pressures  on  the  ver- 
tical faces  ||  to  paper  have  no  vertical  com- 
ponents, and  hence  are  not  drawn.)     From  FIG.  453. 
(2)  we  have 


P  =P0  + 


(3) 


(hy,  being  the  weight  of  a  column  of  homogeneous  fluid  of  unity 
cross-section  and  height  A,  would  be  the  total  pressure  on  the 
base  of  such  a  column,  if  at  rest  and  with  no  pressure  on  the 
upper  base,  and  hence  might  be  called  intensity  due  to  iveigfit.) 
Secondly,  cutting  plane  oblique. — Fig.  453.  Consider  free 
an  infinitely  small  right  triangular  prism  bed,  whose  bases  are 

||  to  the. paper,  while  the  three  side 
faces  (rectangles),  having  areas  =  dF, 
dF^ ,  and  dF^ ,  are  respectively  hori- 
zontal, vertical,  and  oblique ;  let  angle 
cbd  =  a.  The  surface  be  is  a  portion 
of  the  plane  BC  of  Fig.  452.  Given 
p  (=  intensity  of  pressure  on  dF)  and 
a,  required^,  the  intensity  of  pressure 
on  the  oblique  face  bd,  of  area  dJF1^ 
[N.  B. — The  prism  is  taken  very  small 
in  order  that  the  intensity  of  pressure  may  be  considered  con- 
stant over  any  one  face ;  and  also  that  the  weight  of  the  prism 
may  be  neglected,  since  it  involves  the  volume  (three  dimen- 


522  MECHANICS    OF   ENGINEERING. 

sions)  of  the  prism,  while  the  total  face  pressures  involve  only 
two,  and  is  hence  a  differential  of  a  higher  order.] 
From  2  (vert,  compons.)  =  0  we  shall  have 


pdF  =  0  ;  but  dF  '--f-  dFt  =  cos  a  ; 


which  is  independent  of  the  angle  a. 

Hence,  the  intensity  of  fluid  pressure  at  a  given  point  is 
the  same  on  all  imaginary  cutting  planes  containing  the 
point.  This  is  the  most  important  property  of  a  fluid,  and  is 
true  whether  the  liquid  is  at  rest  or  has  any  kind  of  motion  ; 
for,  in  case  of  rectilinear  accelerated  motion,  e.g.,  although  the 
sum  of  the  force-components  in  the  direction  of  the  accelera- 
tion does  not  in  general  =  0,  but  —  mass  X  ace.,  still,  the 
mass  of  the  body  in  question  is  =  weight  -v-  <?,  and  therefore 
the  term  mass  X  ace.  is  a  differential  of  a  higher  order  than 
the  other  terms  of  the  equation,  and  hence  the  same  result 
follows  as  when  there  is  no  motion  (or  uniform  rectilinear 
motion). 

413.  The  Intensity  of  Pressure  is  Equal  at  all  Points  of  any 
Horizontal  Plane  in  a  body  of  homogeneous  fluid  at  rest.  If 
we  consider  a  right  prism  of  the  fluid  in  Fig.  451,  of  small 
vertical  thickness,  its  axis  lying  in  any  horizontal  plane  BC, 
its  bases  will  be  vertical  and  of  equal  area  dF.  The  pressures 
on  its  sides,  being  normal  to  them,  and  hence  to  the  axis,  have 
no  components  ||  to  the  axis.  The  weight  of  the  prism  also 
has  no  horizontal  component.  Hence  from  2  (hor.  comps. 
||  to  axis)  =  0,  we  have,  pl  and  p3  being  the  pressure-intensi- 
ties at  the  two  bases, 

P,dF-p3dF=0;    .:p=p.,    ....    (1) 

which  proves  the  statement  at  the  head  of  this  article. 

It  is  now  plain,  from  this  and  the  preceding  article,  that 
the  pressure-intensity  p  at  any  point  in  a  homogeneous  fluid 
at  rest  is  equal  to  that  at  any  higher  point,  plus  the  weight 


FLUID    PRESSURE. 


523 


of  a  column  of  the  fluid  of  section  unity  and  of  altitude 
(A)  —  vertical  distance  between  the  points. 


(2) 


whether  they  are  in  the  same  vertical  or  not,  and  whatever  be 

the  shape   of  the   containing 

vessel  (or  pipes),  provided  the 

fluid  is   continuous   between 

the  two  points;  for,  Fig.  454, 

by   considering    a    series    of 

small  prisms,  alternately  ver- 

tical and  horizontal,  obcde,  we 


know  that 


Fia.454. 


Pa 


hence,  finally,  by  addition  we  have 


(in  which  h  =  k1  —  A2).  " 

If,  therefore,  upon  a  small  piston  at  0,  of  area  =  FQ  ,  a  force 
P0  be  exerted,  and  an  inelastic  fluid  (liquid)  completely  fills  the 
vessel,  then,  for  equilibrium,  the  force  to  be  exerted  upon  the  pis- 
ton at  e,  viz.,  Pe  ,  is  thus  computed  :  For  equilibrium  of  fluid 
pe  =p0  -f  hy  ;  and  for  equil.  of  piston  <?,  pv  =  P0  -h  F0  ;  also, 


(3) 


From  (3)  we  learn  that  if  the  pistons  are  at  the  same  level 
(h  =  0)  the  total  pressures  on  their  inner  faces  are  directly 
proportional  to  their  areas. 

If  the  fluid  is  gaseous  (2)  and  (3)  are  practically  correct  if 
h  is  not  >  100  feet  (for,  gas  being  compressible,  the  lower 
strata  are  generally  more  dense  than  the  upper),  but  in  (3)  the 
pistons  must  be  fixed,  and  Pe  and  PQ  refer  solely  to  the  in- 
terior pressures. 


524  MECHANICS    OF   ENGINEERING. 

Again,  if  A  is  small  or  p0  very  great,  the  term  hy  may  be 
omitted  altogether  in  eqs.  (2)  and  (3)  (especially  with  gases, 
since  for  them  y  (heaviness)  is  usually  small),  and  we  then 
have,  from  (2), 

P=P.\    •     •     • (4) 

being  the  algebraic  form  of  the  statement:  A  lody  of  fluid 
at  rest  transmits  pressure  with  equal  intensity  in  every  direc- 
tion and  to  all  of  its  parts.  [Principle  of  "Equal  Transmis- 
sion of  Pressure."] 

414.  Moving  Pistons. — If  the  fluid  in  Fig.  454  is  inelastic 
and  the  vessel  walls  rigid,  the  motion  of  one  piston  (o)  through 
a  distance  s0  causes  the  other  to  move  through  a  distance  se  de- 
termined by  the  relation  F0s0  —  Fese  (since  the  volumes  de- 
scribed by  them  must  be  equal,  as  liquids  are  incompressible) ; 
but  on  account  of  the  inertia  of  the  liquid,  and  friction  on  the 
vessel  walls,  equations  (2)  and  (3)  no  longer  hold  exactly,  still 
are  approximately  true  if  the  motion  is  very  slow  and  the 
vessel  short,  as  with  the  cylinder  of  a  water-pressure  engine. 

But  if  the  fluid  is  compressible  and  elastic  (gases  and  vapors ; 
steam,  or  air)  and  hence  of  small  density,  the  effect  of  inertia 
and  friction  is  not  appreciable  in  short  wide  vessels  like  the 
cylinders  of  steam-  and  air-engines,  and  those  of  air-compres- 
sors ;  and  eqs.  (2)  and  (3)  still  hold,  practically,  even  with  high 
piston-speeds.  For  exam  pie,  in  the  space  AB, 
Fig.  455,  between  the  piston  and  cylinder-head 
of  a  steam-engine  (piston  moving  toward  the 
right)  the  intensity  of  pressure,  J9,  of  the 
steam  against  the  moving  piston  B  is  prac- 


FIG.  455.          tically  equal  to  that  against  the  cylinder-head 
A  at  the  same  instant. 

415.  An  Important  Distinction  between  gases  and  liquids 
(i.e.,  between  elastic  and  inelastic  fluids)  consists  in  this : 

A  liquid  can  exert  pressure  against  the  walls  of  the  contain- 
ing vessel  only  by  its  weight,  or  (when  contined  on  all  sides) 
by  transmitted  pressure  coming  from  without  (due  to  piston 
pressure,  atmospheric  pressure,  etc.);  whereas — 


FLUID    PRESSURE.  525 

A  gas,  confined,  as  it  must  be,  on  all  sides  to  prevent  dif- 
fusion, exerts  pressure  on  the  vessel  not  only  by  its  weight, 
but  by  its  elasticity  or  tendency  to  expand.  If  pressure  from 
without  is  also  applied,  the  gas  is  compressed  and  Exerts  a  still 
greater  pressure  on  the  vessel  walls. 

\     I 
416.  Component,   of   Pressure,   in    a  Given    Direction, — Let 

ABCD,  whose  area  =  dF,  be  a  small  element  of  \a  surface, 

plane  or  curved,  and  j9  the  intensity  of 

fluid  pressure  upon  this  element,  then 

the  total  pressure  upon  it  \$>pdF,  and  is 

of  course  normal  to  it.     Let  A' B'  CD  be 

the  projection  of  the  element  dF  upon     c\^        /\\fe--''* 

a  plane  CDM.  making  an  angle  a  with 

the  element,  and  let  it  be  required  to 

find  the  value  of  the  component  of  pdF 

in  a  direction  normal  to  this  last  plane  (the  other  component 

being  understood  to  be  ||  to  the  same  plane).     We  shall  have  < 


Compon.  ofpdF  ~\  to  CDM  =  pdFcos  a  =  p(dFiOQ).  (1)"' 


But  dF  .  cos  a  =  area  A'B'CD,  the  projection  of 
the  plane  CDM. 

.*.  Compon.  "1  to  plane  CDM  =p  X  (project.  ofdFon  CDM]\ 

i.e.,  the  component  of  fluid  pressure  (on  an  element  of  a  sur- 
face) in  a  given  direction  (the  other  component  being  "I  to 
the  first)  is  found  ly  multiplying  the  intensity  of  the  pressure 
ty  the  area  of  the  projection  of  the  element  upon  a  plane  ~\  to 
the  given  direction,  v 

It  is  seen,  as  an  example  of  this,  that  if  the  fluid  pressures 
on  the  elements  of  the  inner  surface  of  one  hemisphere  of  a 
hollow  sphere  containing  a  gas  are  resolved  into  components  1 
and  ||  to  the  plane  of  the  circular  base  of  the  hemisphere,  the 
sum  of  the  former  components  simply  —  7rr*p,  where  r  is  the 
radius  of  the  sphere,  and^>  the  intensity  of  the  fluid  pressure  ; 
for,  from  the  foregoing,  the  sum  of  these  components  is  just 
the  same  as  the  total  pressure  would  be,  having  an  intensity  p> 


526 


MECHANICS   OF   ENGINEERING. 


on  a  great  circle  of  the  sphere,  the  area,  m?,  of  this  circle  being 
the  sum  of  the  areas  of  the  projections,  upon  this  circle  as  a 
base,  of  all  the  elements  of  the  hemispherical  surface.  (Weight 
of  fluid  neglected.) 

A  similar  statement  may  be  made  as  to  the  pressures  on 

e  inner  curved  surface  of  a  right  cylinder. 


417.  Non-planar  Pistons. — From  the  foregoing  it  follows  that 
I/  >fehe  sum  of  the  components  ||  to  the  piston-rod,  of  the  fluid 
i/  pressures  upon  the  piston  at  A,  Fig.  457,  is  just  the  same  as  at 
B,  if  the  cylinders  are  of  equal  size  and  the  steam,  or  air,  is  at 
the  same  tension.  For  the  sum  of  the  projections  of  all  the 
elements  of  the  curved  surface  of  A  upon  a  plane  ~|  to  the 
piston-rod  is  always  =:  nr*  =  area  of  section  of  cylinder-bore. 


SP 

|^!®:.v5 

S 

If  the  surface  of  A  is  symmetrical  about  the  axis  of  the  cylin- 
der the  other  components  (i.e.,  those  ~|  to  the  piston-rod)  will 
neutralize  each  other.  If  the  line  of  intersection  of  that  sur- 
face with  the  surface  of  the  cylinder  is  not  symmetrical  about 
the  axis  of  the  cylinder,  the  piston  may  be  pressed  laterally 
against  the  cylinder-wall,  but  the  thrust  along  the  rod  or 
"  working  force"1  (§  128)  is  the  same  (except  for  friction  in- 
duced by  the  lateral  pressure),  in  all  instances,  as  if  the  surface 
were  plane  and  ~]  to  piston-rod. 

418.  Bramah,  or  Hydraulic,  Press,— This  is  a  familiar  instance 
of  the  principle  of  transmission  of  fluid  pressure.  Fig.  458. 
Let  the  small  piston  at  0  have  a  diameter  d  =  1  inch  =  -^  ft., 
while  the  plunger  E,  or  large  piston,  has  a  diameter  d'  =  AE 
=  CD  =  15  in.  =  £  ft.  The  lever  MJV  weighs  G,  =  3  IDS., 
and  a  weight  G  =  40  Ibs.  is  hung  at  M.  The  lever-arms  of 
these  forces  about  the  fulcrum  N  are  given  in  the  figure. 
The  apparatus  being  full  of  water  (oil  is  often  used),  the  fluid 
pressure  P0  against  the  small  piston  is  found  by  putting 


FLUID    PRESSURE. 


52T 


^(moms.  about  2V)  =  0  for  the   equilibrium  of   the  lever; 
whence  [ft.,  lb.,  sec.] 

po  x  i  _  40  X  3  -  3  X  2  =  0.     /.  P0  =  126  Ibs.  * 


FIG.  458. 


But,  denoting  atmospheric  pressure  by  pa,  and  that  of  the 
water  against  the  piston  by  p0  (per  unit  area),  we  may  also 
write 


Solving  for  p0  ,  we   have,  putting  pa  —  14.7  X  144  Ibs.  per 
eq.  ft, 


r 
1 


0  =     126  -f-      (^M  +  14.7  X  144  =  25236  Ibs.  per  sq.  ft. 


Hence  at  e  the  press,  per  unit  area,  from  §  409,  and  (2),  §  413,  is 
pe  =pn  -f-  hy  =  25236  +  3  X  62.5  =  25423  Ibs.  per  sq.  ft. 

=  175.6  Ibs.  per  sq.  inch  or  11.9  atmospheres,  and  the  total 
upward  pressure  at  e  on  base  of  plunger  is 

n  ^pe  =  ±  7t(*-y  x  25423  =  31194  Ibs, 

or  almost  16  tons  (of  2000  Ibs.  each).  The  co  repressive  force 
upon  the  block  or  bale,  C,  =  P  less  the  weight  of  the  plunger 
and  total  atmos.  pressure  on  a  circle  of  15  in.  diameter. 


628 


MECHANICS   OF  ENGINEEKING. 


419.  The  Dividing  Surface  of  Two  Fluids  (which  do  not  mix)  in 
Contact,  and  at  Rest,  is  a  Horizontal  Plane. — For,  Fig.  459,  sup- 
posing any  two  points  e  and  O  of  this  sur- 
face to  be  at  different  levels  (the  pressure 
at  0  being  _£>0,  that  at  epe,  and  the  heavi- 
nesses of  the  two  fluids  yl  and  y^  respec- 
tively), we  would  have,  from  a  considera- 
tion of  the  two  elementary  prisms  eb  and 
bO  (vertical  and  horizontal),  the  relation 


. 


FIG.  459.- 


while  from  the  prisms  ec  and  cO,  the  relation 


These  equations  are  conflicting,  hence  the  aoove  supposition 
is  absurd.     Therefore  the  proposition  is  true. 

For  stable  equilibrium,  evidently,  the  heavier  fluid  must  oc- 
cupy the  lowest  position  in  the  vessel,  and  if  there  are  several 
fluids  (which  do  not  mix),  they  will  arrange 
themselves  vertically,  in  the  order  of  their  den- 
sities, the  heaviest  at  the  bottom,  Fig.  460.  On 
account  of  the  property  called  diffusion  the  par- 
ticles of  two  gases  placed  in  contact  soon  inter- 
mingle and  form  a  uniform  mixture.  This  fact 
gives  strong  support  to  the  "  Kinetic  Theory  of 
Gases"  (§  408). 


FIG.  460. 


420.  Free  Surface  of  a  Liquid  at  Rest. — The  surface  (of  a 
liquid)  not  in  contact  with  the  walls  of  the  containing  vessel 
is  called  a  free  surface,  and  is  necessarily 
horizontal  (from  §  419)  when  the  liquid  is  at 
rest.  Fig.  461.  (A  gas,  from  its  tendency 
to  indefinite  expansion,  is  incapable  of  hav- 
ing a  free  surface.)  This  is  true  even  if  the 
space  above  the  liquid  is  vacuous,  for  if  the 
surface  were  inclined  or  curved,  points  in  the 
body  of  the  liquid  and  in  the  same  horizon- 
tal plane  would  have  different  heights  (or  "  heads")  of  liquid 


FIG.  461. 


TWO   LIQUIDS   IN  BENT   TUBE. 


529 


between  them  and  the  surface,  producing  different  intensities 
of  pressure  in  the  plane,  which  is  contrary  to  §  413. 

When  large  bodies  of  liquid  like  the  ocean  are  considered, 
gravity  can  no  longer  be  regarded  as  acting  in  parallel  lines ; 
consequently  the  free  surface  of  the  liquid  is  curved,  being  ~| 
to  the  direction  of  (apparent)  gravity  at  all  points.  For  ordi- 
nary engineering  purposes  (except  in  Geodesy)  the  free  surface 
of  water  at  rest  is  a  horizontal  plane. 

421.  Two  Liquids  (which  do  not  mix)  at  Rest  in  a  Bent  Tube 
open  at  Both  Ends  to  the  Air,  Fig.  460 ;  water  and  mercury,  for 
instance.  Let  their  heavinesses  be  Yl 
and  YI  respectively.  The  pressure  at  e 
may  be  written  (§  413)  either 


or 


according  as  we  refer  it  to  the  water 
column  or  the  mercury  column  and 
their  respective  free  surfaces  where  the 
pressure  JPO,  =^oa  =  pa  —  atmos.  press. 
e  is  the  surface  of  contact  of  the  two  liquids.  Hence  we  have 


i.e.,  the  heights  of  the  free  surfaces  of  the  two  liquids  above  the 
surface  of  contact  are  inversely  proportional  to  their  respec- 
tive heavinesses. 

EXAMPLE.  —  If  the  pressure  at  e  =  2  atmospheres  (§  396)  we 
shall  have  from  (1)  (inch-lb.-sec.  system  of  units) 

h^  =pe—pa  =  2x  14.7  —  14.7  —  14.7  Ibs.  per  sq.  inch. 
.-.  A2  must  =  14.7  -r-  [848.7  +•  1728]  =  30  inches 

(since,  for  mercury,  yz  =  848.7  Ibs.  per  cub.  ft.).  Hence, 
from  (3), 


KY*        30  X  [848.7  -*-  1728] 


530  MECHANICS    OF   ENGINEERING. 

i.e.,  for  equilibrium,  and  that pe  may  =  2  atmospheres,  ht  and 
Aa  (of  mercury  and  water)  must  be  30  in.  and  34  feet  respec- 
tively. 

422.  City  Water-pipes. — If  h  =  vertical  distance  of  a  point 
B  of  a  water-pipe  below  the  free  surface  of  reservoir,  and  the 
water  be  at  rest,  the  pressure  on  the  inner  surface  of  the  pipe 
at  B  (per  unit  of  area)  is 

p  =p0  +  hy ;   and  here  j?0  =pa  =  atmos.  press. 
EXAMPLE. — If  h  =  200  ft.  (using  the  inch,  lb.,  and  second) 
p  =  14.7  +  [200  X  12] [62.5  +  1728]  =  101.5  Ibs.  per  sq.  in. 

The  term  hy,  alone,  =  86.8  Ibs.  per  sq.  inch,  is  spoken  of  as  the 
hydrostatic  pressure  due  to  200  feet  height,  or  *\Head,"  of 
water.  (See  Trautwine's  Pocket  Book  for  a  table  of  hydro- 
static pressures  for  various  depths.) 

If,  however,  the  water  \&  flowing  through  the  pipe,  the  pres- 
sure against  the  interior  wall  becomes  less  (a  problem  of  Hy- 
drodynamics to  be  treated  subsequently),  while  if  that  motion 
is  suddenly  checked,  the  pressure  becomes  momentarily  much 
greater  than  the  hydrostatic.  This  shock  is  called  "water- 
ram"  and  "  water-hammer,"  and  may  be  as  great  as  200  to  300 
Ibs.  per  sq.  inch. 

423.  Barometers  and  Manometers  for  Fluid  Pressure. — If   a 
tube,  closed  at  one  end,  is  filled  with  water,  and  the  other  ex- 
tremity is  temporarily  stopped  and  afterwards 
opened  under  water,  the  closed  end  being  then 
a  (vertical)  height  =  h  above  the  surface  of 
the  water,  it  is  required  to  find  the  intensity, 
p0 ,  of  fluid  pressure  at  the  top  of  the  tube,  sup- 
posing it  to  remain   filled  with  water.     Fig. 
463.     At  E  inside  the   tube  the  pressure  is 
14.7  Ibs.  per  sq.  inch,  the  same  as  that  outside 
at  the  same  level  (§  413) ;  hence,  from  pE  =  p0 


BAROMETERS.  531 

EXAMPLE. — Let  h  =  10  feet  (with  inch-lb.-sec.  system) ;  then 
p0  =  14.7  —  120  X  [62.5  -j-  1728]  =  10.4  Ibs.  per  sq.  inch, 

or  about  §  of  an  atmosphere.  If  now  we  inquire  the  value 
of  h  to  make p0  =  zero,  we  putpE  —  hy  =  0  and  obtain  h  = 
408  inches,  =  34  ft.,  which  is  called  the  height  of  the  water- 
barometer.  Hence,  Fig.  463$,  ordinary  atmospheric  pressure 
will  not  sustain  a  column  of  water  higher  than  34  feet.  If 
mercury  is  used  instead  of  water  the  height  supported  by  one 
atmosphere  is 

I  =  14.7  -=-  [848.7  -f-  1728]  =  30  inches, 

=  76  centims.  (about),  and  the  tube  is  of  more  manageable 
proportions  than  with  water,  aside  from  the  ad- 
vantage that  no  vapor  of  mercury  forms  above 
the  liquid  at  ordinary  temperatures  [In  fact,  the 
water-barometer  height  b  =  34  feet  has  only  a 
theoretical  existence  since  at  ordinary  tempera- 
tures (40°  to  80°  Fahr.)  vapor  of  water  would 
form  above  the  column  and  depress  it  by  from 
0.30  to  1.09  ft.].  Such  an  apparatus  is  called  a 
Barometer ',  and  is  used  not  only  for  measuring 
the  varying  tension  of  the  atmosphere  (from  14.5 
to  15  Ibs.  per  sq.  inch,  according  to  the  weather  and  height 
above  sea-level),  but  also  that  of  any  body  of  gas.  Thus,  Fig. 
464,  the  gas  in  D  is  put  in  communication  with 
the  space  above  the  mercury  in  the  cistern  at 
(7;  and  we  have  p  —  hy,  where  y  =  heav.  of 
mercury,  and  JP  is  the  pressure  on  the  liquid  in 
the  cistern.  For  delicate  measurements  an  at- 
tached thermometer  is  also  used,  as  the  heavi- 
ness Y  varies  slightly  with  the  temperature. 

If  the  vertical  distance  CD  is  small,  the  ten- 
sion in  C  is  considered  the  same  as  in  D. 

For  gas-tensions  greater  than  one  atmosphere, 
the  tube  may  be  left  open  at  the  top,  forming  an  open  ma- 


FIG.  463a. 


FIG.  464. 


532 


MECHANICS    OF    ENGINEERING. 


nometer,  Fig.  465.     In  this  case,  the  tension  of  the  gas  above 
the  mercury  in  the  cistern  is 


(i) 


FIG.  465. 


in  which  ~b  is  the  height  of  mercury  (about  30 
in.)  to  which  the  tension  of  the  atmosphere  above 
the  mercury  column  is  equivalent. 

EXAMPLE. — If  h  —  51   inches,   Fig.   465,   we 
have  (ft.,  lb.,  sec.) 


p  =  [4.25  ft.  +  2.5  ft.]  848.7  =  5728  Ibs.  per  sq.  foot 
=  39.7  Ibs.  per  sq.  inch  =  2.7  atmospheres. 

Another  form  of  the  open  manometer  consists  of  a  U  tube, 
Fig.  464,  the  atmosphere  having  access  to  one  branch,  the  gas 
to   be  examined,  to  the  other,  while  the 
mercury  lies  in  the  curve.     As  before,  we 
have 

(2) 


tn 


where  pa  =  atmos.  tension,  and  b  as  above. 
The  tension  of  a  gas  is  sometimes  spoken 
of  as  measured  by  so  many  inches  of  mer- 
cury. For  example,  a  tension  of  22.05  FIG.  466. 
Ibs.  .per  sq.  inch  (1J  atmos.)  is  measured  by  45  inches  of  mer- 
cury -in  a  vacuum  manometer  (i.e.,  a  common  barometer), 
Fig.  464.  "With  the  open  manometer  this  tension  (1£  atmos.) 
would  be  indicated  by  15  inches  of  actual  mercury,  Figs.  465 
and  466.  An  ordinary  steam-gauge  indicates  the  'excess  of 
tension  over  one  atmosphere ;  thus  "  40  Ibs.  of  steam"  implies 
a  tension  of  40  +  14.7  =  54.7  Ibs.  per  sq.  in. 

The  Bourdon  steam-gauge  in  common  use  consists  of  a 
curved  elastic  metal  tube  of  flattened  or  elliptical  section 
(with  the  long  axis  ~]  to  the  plane  of  the  tube),  and  has  one 
end  fixed.  The  movement  of  the  other  end,  which  is  free  and 


TENSION    OF   GASES. 


533 


closed,  by  proper  mechanical  connection  gives  motion  to  the 
pointer  of  a  dial.  This  movement  is  caused  by  any  change  of 
tension  in  the  steam  or  gas  admitted,  through  the  fixed  end,  to 

O  *  O 

the  interior  of  the  tube.  As  the  tension  increases  the  ellip- 
tical section  becomes  less  flat,  i.e.,  more  nearly  circular,  caus- 
ing the  two  ends  of  the  tube  to  separate  more,  widely,  i.e.,  the 
free  end  moves  away  from  the  fixed  end  ;  and  vice  versa. 

Such  gauges,  however,  are  not  always  reliable.  They  are 
graduated  by  comparison  with  mercury  manometers;  and 
should  be  tested  from  time  to  time  in  the  same  way. 

424.  Tension  of  Illuminating  Gas. — This  is  often  spoken  of  as 
measured  by  inches  of  water  (from  1  to  3  inches  usually). 
Strictly  it  should  be  stated  that  this 
water-height  measures  the  excess  of 
its  tension  over  that  of  the  atmos- 
phere. Thus,  in  Fig.  466,  water 
being  used  instead  of  mercury,  h  = 
say  2  inches,  while  b  =  408  inches. 

This  difference  of  tension  may  be 
largely  affected  by  a  change  in  the 
barometer  due  to  the  weather/ or  by 
a  difference  in  altitude,  as  the  follow- 
ing example  will  illustrate : 

EXAMPLE. — Supposing  the  gas  at  rest,  and  the  tension  at  the 
gasometer  A,  Fig.  467,  to  be  "two  inches  of  water,"  required 
the  water-column  h"  (in  open  tube)  that  the  gas  will  support 
in  the  pipe  at  B,  120  feet  (vertically)  above  the  gasometer. 
Let  the  temperature  be  freezing  (nearly),  and  the  outside  air  at 
a  tension  of  14.7  Ibs.  per  sq.  inch ;  the  heaviness  of  the  gas  at 
this  temperature  being  0.036  Ibs.  per  cubic  foot.  For  the 
small  difference  of  120  ft.  we  may  treat  both  the  atmosphere 
and  the  gas  as  liquids,  that  is,  of  constant  density  throughout 
the  vertical  column,  and  therefore  apply  the  principles  of 
§  413  ;  with  the  following  result : 

The  tension  of  the  outside  air  at  JSy  supposed  to  be  at  the 
same  temperature  as  at  A,  will  sustain  a  water-column  less 
than  the  408  inches  at  A  by  an  amount  corresponding  to  the 


FIG.  467. 


534  MECHANICS    OF    ENGINEERING. 

120  feet  of  air  between,  of  the  heaviness  .0807  Ibs.  per  cub. 
ft.  120  feet  of  air  weighing  .0807  Ibs.  per  cub.  ft.  will  balance 
0.154  ft.  of  water  weighing  62.5  Ibs.  per  cubic  ft.,  i.e.,  1.85 
inches  of  water.  Now  the  tension  of  the  gas  at  B  is  also  less 
than  its  tension  at  J.,  but  the  difference  is  not  so  great  as  with 
the  outside  air,  for  the  120  ft.  of  gas  is  lighter  than  the  120  ft. 
of  air.  Since  120  ft.  of  gas  weighing  0.036  Ibs.  per  cubic  ft. 
will  balance  0.0691  ft.,  or  0.83  inches,  of  water,  therefore  the 
difference  between  the  tensions  of  the  two  fluids  at  B  is  greater 
than  at  A  by  (1.85  —  0.83  =  )  1.02. inches;  or,  at  B  the  total 
difference  is  2.00  +  1.02  =  3.02  inches.v 

Hence  if  a  small  aperture  is  made  in  the  pipe  at  B  the  gas 
will  flow  out  with  greater  velocity  than  at  A.  At  Ithaca, 
!N.  Y.,  where  the  University  buildings  are  400  ft.  above  the 
gas-works,  this  phenomenon  is  very  marked. 

"When  the  difference  of  level  is  great  the  decrease  of  tension 
as  we  proceed  upward  in  the  atmosphere,  even  with  constant 
temperature,  does  not  follow  the  simple  law  of  §413;  see 
§477. 

For  velocity  of  flow  of  gases  through  orifices,  see  §  548,  etc. 

425.  Safety-valves. — Fig.  468.  Kequired  the  proper  weight 
G  to  be  hung  at  the  extremity  of  the  horizontal  lever  ABY 

with  fulcrum  at  B,  that  the  flat 
disk- valve  E  shall  not  be  forced 
upward  by  the  steam  pressure,  p' , 
until  the  latter  reaches  a  given 
value  —  p.  Let  the  weight  of 
the  arm  be  6r, ,  its  centre  of  grav- 
ity being  at  £7,  a  distance  =  o 

from  B ;  the  other  horizontal  distances  are  marked  in  the 
figure. 

1  Suppose  the  valve  on  the  point  of  rising ;  then  the  forces 
acting  on  the  lever  are  the  fulcrum-reaction  at  B,  the  weights 
G  and  Gl ,  and  the  two  fluid-pressures  on  the  disk,  viz. :  Fpa 
(atmospheric)  downward,  and  Fp  (steam)  upward.  Hence, 
from  ^(moms.  B)  =  0, 

Gb  +  G,c  +  Fpaa  -  Fpa  =  0.      ...    (1) 


BURSTING   OF  PIPES. 


535 


•Solving,  we  have 


(2) 


EXAMPLE.—  With  a  =  2  inches,  £  =  2  feet,  c  =  1  foot 
Gl  =  4  Ibs.,j9  =  6  atmos.,  and  diam.  of  disk  =  1  inch;  with 
the  foot  and  pound, 


X  m  X  144  -  1  X  14.7  X  144]  -4 
.-.  G  =  2.81  Ibs. 


[Notice  the  cancelling  of  the  144;  for  F(p  —p^)  is  pounds, 
being  one  dimension  of  force,  if  the  pound  is  selected  as  the 
unit  of  force,  whether  the  inch  or  foot  is  used  in  both  fac- 
tors.] Hence  when  the  steam  pressure  has  risen  to  6  atmos. 
(=  88.2  Ibs.  per  square  inch)  (corresponding  to  73.5  Ibs.  persq. 
in.  by  steam-gauge)  the  valve  will  open  if  G  =  2.81  Ibs.,  or  be 
on  the  point  of  opening. 

426.  Proper  Thickness  of  Thin  Hollow  Cylinders  (i.e,,  Pipes 
and  Tubes)  to  Resist  Bursting  by  Fluid  Pressure. 

CASE  I.  /Stresses  in  the  cross-section  due  to  End  Pressure; 
Fig.  469. — Let  AB  be  the  circular  cap  clos- 
ing the  end  of  a  cylindrical  tube  containing 
fluid   at   a   tension  =  p.      Let  r  =  internal 
radius  of  the  tube  or  pipe.    Then  considering 
the  cap  free,  neglecting  its  weight,  we  have 
three  sets  of    ||   forces  in  equilibrium   (see 
II  in  figure),  viz.:    the  iuternal  fluid  pres- 
sure =  7tr*p-,    the    external   fluid    pressure 
=  nr*pa ;  while  the  total  stress  (tensile)  on 
the  small  ring,  whose  area  now  exposed  is 
2 nrt  (nearly),  is  =  <^nrtpl ,  where  t  is  the  thickness  of  the  pipe, 
and  j^  the  tensile  stress  per  unit  area  induced  by  the  end-pres- 
sures (fluid). 


636  MECHANICS    OF    ENGINEERING. 

For  equilibrium,  therefore,  we  may  put  ^(hor.  comps.)  =  0 ; 

»i  =  0; 

.....    (1) 


_  r(p  -pa) 
••Pi ^ • 


(Strictly,  the  two  circular  areas  sustaining  the  fluid  pressures 
are  different  in  area,  but  to  consider  them  equal  occasions  but 
a  small  error.) 

Eq.  (1)  also  gives  the  tension  in  the  central  section  of  a  thin 
hollow  sphere,  under  bursting  pressure. 

CASE  II.  Stresses  in  the  longitudinal  section  of  pipe,  due  to 
radial  fluid  pressures.* — Consider  free  the  half  (semi-circular) 

of  any  length  I  of  the  pipe,  be- 
tween two  cross-sections.  Take  an 
axis  X  (as  in  Fig.  470)  "|  to  the 
longitudinal  section  which  has  been 
made.  Let  pz  denote  the  tensile 
stress  (per  unit  area)  produced  in 
the  narrow  rectangles  exposed  at  A 
and  B  (those  in  the  half-ring  edges, 
having  no  X  components,  are  not 
drawn  in  the  figure).  On  the  in- 
ternal curved  surface  the  fluid  pres- 
sure is  considered  of  equal  intensity 
=  ^>  at  all  points  (practically  true  even  with  liquids,  if  2r  is 
small  compared  with  the  head  of  water  producing  p).  The 
fluid  pressure  on  any  dF  or  elementary  area  of  the  internal 
curved  surface  is  =  pdF.  Its  X  component  (see  §  416)  is 
obtained  by  multiplying^?  by  the  projection  of  dF  on  the  ver- 
tical plane  ABC,  and  since  p  is  the  same  for  all  the  dF'&  of 
the  curved  surface,  the  sum  of  all  the  X components  of  the  in- 
ternal fluid  pressures  must  =  p  multiplied  by  the  area  of  rect- 
angle ABCD,  =  %rlp  ;  and  similarly  the  X  components  of  the 


FIG.  470. 


*  Analytically  this  problem  is  identical  with  that  of  the  smooth  cord  on 
a  smooth  cylinder,  §  169,  and  is  seen  to  give  the  same  result. 


BURSTING   OF  PIPES. 


v 

external  atmos.  pressures  =  %rlpa  (nearly).  The  tensile  stresses 
(  ||  to  X)  are  equal  to  2%3  ;  hence  for  equilibrium,  ^2X—  0 
gives 

3  —  %rlp  +  2rlpa  =  0  ; 


This  tensile  stress,  called  hoop  tension,  p^,  opposing  rupture  by 
longitudinal  tearing,  is  seen  to  be  double  the  tensile  stress^ 
induced,  under  the  same  circumstances,  on  the  annular  cross' 
section  in  Case  I.  Hence  eq.  (2),  and  not  eq.  (1),  should  be 
used  to  determine  a  safe  value  for  the  thickness  of  metal,  £,  or 
any  other  one  unknown  quantity  involved  in  the  equation. 

For  safety  against  rupture,  we  must  put  pz  —  T'  ',  a  safe 
tensile  stress  per  unit  area  for  the  material  of  the  pipe  or  tube 
(see  §§  195  and  203)  ; 


(For  a  thin  hollow  sphere,  t  may  be  computed  from  eq.  (1)  ; 
that  is,  need  be  only  half  as  great  as  with  the  cylinder,  other 
things  being  equal.) 

EXAMPLE.  —  A  pipe  of  twenty  inches  internal  diameter  is  to 
contain  water  at  rest  under  a  head  of  340  feet  ;  required  the 
proper  thickness,  if  of  cast-iron. 

340  feet  of  water  measures  10  atmospheres,  so  that  the  in 
ternal  fluid  pressure  is  11  atmospheres  ;  but  the  external  pres- 
sure pa  being  one  atmos.,  we  must  write  (inch,  lb.,  sec.) 

(p—pa)  =  10  X  14.  Y  =  147.0  Ibs.  per  sq.  in.,  and  r  =  10  in., 

while  (§  203)  we  may  put  T'  =  J  of  9000  =  4500  Ibs.  per  sq. 
in.  ;  whence 


t  =      ,  =  0.326  inches. 

4500 


538  MECHANICS   OF   ENGINEERING. 

But  to  insure  safety  in  handling  pipes  and  imperviousness  to 
the  .water,  a  somewhat  greater  thickness  is  adopted  in  practice 
than  given  by  the  above  theory. 

Thus,  Weisbach  recommends  (as  proved  experimentally  also) 
for 


Pipes  of  sheet  iron,  t  =  [0.00172  rA  +  0.12" 

"0.00476  rA  +  0.34T 
U00296  rA  +  0.16= 
"0.01014  rA  +  0.21 
"0.00484  rA  +  0.16= 


u        .. 


cast  t  = 

copper  t  = 

"      "  lead  t  = 

I         "      "  zinc  t  = 


inches ; 


in  which  t  =  thickness  in  inches,  r  =  radius  in  inches,  and  A 
=  excess  of  internal  over  external  fluid  pressure  (i.e.,  p  —  pa) 
expressed  in  atmospheres. 

For  instance,  for  the  example  just  given,  we  should  have 
(cast-iron) 

t  —  .00476  X  10  X  10  +  0.34  =  0.816  inches. 

If  the  pipe  is  subject  to  "  water-ram"  (§  422)  the  strength 
should  be  much  greater.  To  provide  against  "  water-ram," 
Mr.  J.  T.  Fanning,  on  p.  453  of  his  "  Hydraulic  and  "Water- 
supply  Engineering,"  advises  adding  230  feet  to  the  static 
head  in  computing  the  thickness  of  cast-iron  pipes. 

For  thick  hollow  cylinders  see  Rankine's  Applied  Mechan- 
ics, p.  290,  and  Cotterill's  Applied  Mechanics,  p.  403. 

427.  Collapsing  of  Tubes  under  Fluid  Pressure.  (Cylindrical 
boiler-flues,  for  example.) — If  the  external  exceeds  the  internal 
fluid  pressure,  and  the  thickness  of  metal  is  small  compared 
with  the  diameter,  the  slightest  deformation  of  the  tube  or 
pipe  gives  the  external  pressure  greater  capability  to  produce 
a  further  change  of  form,  and  hence  possibly  a  final  collapse ; 
just  as  with  long  columns  (§  303)  a  slight  bending  gives  great 
advantage  to  the  terminal  forces.  Hence  the  theory  of  §  426 
is  inapplicable.  According  to  Sir  Wm.  Fairbairn's  experi- 
ments (1858)  a  thin  wrought-iron  cylindrical  (circular)  tube 
will  not  collapse  until  the  excess  of  external  over  internal 
pressure  is 


COLLAPSE   OF  TUBES.  539 

j?(in  Ibs.  per  sq.  in.)  =  9672000  1~.     .     .  (1)  .     .   (not  homog.) 

(t,  I,  and  d  must  all  be  expressed  in  the  same  linear  unit.) 
Here  t  =  thickness  of  the  wall  of  the  tube,  d  its  diameter,  and 
Z  its  length  ;  the  ends  being  understood  to  be  so  supported  aa 
to  preclude  a  local  collapse. 

EXAMPLE.—  With  I  =  10  ft.  =  120  inches,  d  =  4  in.,  and  t  = 
J$  inch,  we  have 

p  =  9672000  P-j^-  -5-  (120  X  4)1  =  201.5  Ibs.  per  sq.  inch. 


For  safety,  %  of  this,  viz.  40  Ibs.  per  sq.  inch,  should  not  be 
exceeded  ;  e.g.,  with  14.7  Ibs.  internal  and  54.7  Ibs.  external. 


540  MECHANICS   OF  ENGINEERING. 


CHAPTEE  II. 

HYDROSTATICS  (Continued)— PRESSURE  OF  LIQUIDS  IN  TANKt 
AND  RESERVOIRS. 

428.  Body  of  Liquid  in  Motion,  but  in  Relative  Equilibrium. — 
By  relative  equilibrium  it  is  meant  that  the  particles  are  not 
changing  their  relative  positions,  i.e.,  are  not  moving  among 
each  other.  On  account  of  this  relative  equilibrium  the  fol- 
lowing problems  are  placed  in  the  present  chapter,  instead  of 
under  the  head  of  Hydrodynamics,  where  they  strictly  belong. 
As  relative  equilibrium  is  an  essential  property  of  rigid  bodies, 
we  may  apply  the  equations  of  motion  of  rigid  bodies  to  bodies 
of  liquid  in  relative  equilibrium. 

CASE  I.  All  the  particles  moving  in  parallel  right  lines 
with  equal  velocities  ;  at  any  given  instant  (i.e.,  a  motion  of 
translation.) — If  the  common  velocity  is  constant  we  have  a 
uniform  translation,  and  all  the  forces  acting  on  any  one  par- 
ticle are  balanced,  as  if  it  were  not  moving  at  all  (according  to 
Newton's  Laws,  §  54);  hence  the  relations  of  internal  pressure, 
free  surface,  etc.,  are  the  same  as  if  the  liquid  were  at  rest. 
Thus,  Fig.  471,  if  the  liquid  in  the  moving  tank  is  at  rest  rel- 
v  atively  to  the  tank  at  a  given  instant,  with 
its  free  surface  horizontal,  and  the  motion 
of  the  tank  be  one  of  translation  with  a  uni- 
form velocity,  the  liquid  will  remain  in  this 
condition  of  relative  rest,  as  the  motion 

FIQ.  471. 

proceeds. 

But  if  the  velocity  of  the  tank  is  accelerated  with  a  constant 
acceleration  =p  (this  symbol  must  not  be  confused  with  p 
for  pressure),  the  free  surface  will  begin  to  oscillate,  and  finally 
come  to  relative  equilibrium  at  some  angle  a  with  the  horizon- 
tal, which  is  thus  found,  when  the  motion  is  horizontal.  See 
Fig.  472,  in  which  the  position  and  value  of  of  are  the  same, 
whether  the  motion  is  uniformly  accelerated  from  left  to  right 


RELATIVE   EQUILIBRIUM    OF   LIQUIDS. 


541 


FIG.  472. 


or  uniformly  retarded  from  right  to  left.     Let  0  be  the  lowest 

point  of  the  free  surface,  and  Ob  a  _v    n     *.?  - 

small  prism  of  the  liquid  with  its 
axis  horizontal,  and  of  length  =  % ; 
nb  is  a  vertical  prism  of  length  =  *~T~ 
3,  and  extending  from  the  extremity 
of  Ob  to  the  free  surface.  The 
pressure  at  both  0  and  n  is  pa  = 
atmos.  pres.  Let  the  area  of  cross- 
section  of  both  prisms  be  =  dF. 

Now  since  Ob  is  being  accelerated  in  direction  ^(horizont.), 
the  difference  between  the  forces  on  its  two  ends,  i.e.,  its  ^Xy 
must  =  its  mass  X  accel,  (§  109). 

.\pbdF— padF  =  [xdF.  y-?-g]p.    .    .    .    (1)' 

(y  =  heaviness  of  liquid ;  pb  =  press,  at  b) ;  and  since  the  ver- 
tical prism  nb  has  no  vertical  acceleration,  the  -^(vert.  com- 
pons.)  for  it  must  =  0. 

s.pbdF-padF-zdF.y=Q (2) 

From  (1)  and  (2), 

ffiy  ~ .  £ p  frf\ 

V  "«T 

Hence  On  is  a  right  line,  and  therefore 


tan  a.   or  — ,   =^ 

»       g 


(4) 


[Another,  and  perhaps  more  direct,  method  of  deriving  this 
result  is  to  consider  free  a  small  particle  of  the  liquid  lying  in. 
the  surface.  The  forces  acting  on  this  particle  are  two :  the 
first  its  weight  =  dG  ;  and  the  second  the  resultant  action  of 
its  immediate  neighbor-particles.  Now  this  latter  force  (point- 
ing obliquely  upward)  must  be  normal  to  the  free  surface  of 
the  liquid,  and  therefore  must  make  the  unknown  angle  a  with 
the  vertical.  Since  the  particle  has  at  this  instant  a  rectilinear 
accelerated  motion  in  a  horizontal  direction,  the  resultant  of  the 
two  forces  mentioned  must  be  horizontal  and  have  a  value  = 
mass  X  acceleration.  That  is,  the  diagonal  formed  on  the  two 


542 


MECHANICS   OF   ENGINEERING. 


forces  must  be  horizontal  and  have  the  value  mentioned,  = 
(dG  -r-  g)p ;  while  from  the  nature  of  the  figure  (let  the  stu- 
dent make  the  diagram  for  himself)  it  must  also  =  dG  tan  a. 


-,„  , 
dG  tan  a  =  — . p ;  or,  tan  a  =  •£. 

g  9 


Q.  E.  D.I ' 


If  the  translation  were  vertical,  and  the  acceleration  upward 
[i.e.,  if  the  vessel  had  a  uniformly  accelerated  upward  motion 
or  a  uniformly  retarded  downward  motion],  the  free  surface 
would  be  horizontal,  but  the  pressure  at  a  depth  =  h  below  the 
surface  instead  of  p  =pa  -\-hy  would  be  obtained  as  follows : 
Considering  free  a  small  vertical  prism  of  height  =  h  with 
upper  base  in  the  free  surface,  and  putting  ^(vert.  compons.) 
=  mass  X  acceleration,  we  have 

dF.p  -  dF.pa  -  hdF.  y  =  MF'  y  . p; 


(5) 


If  the  acceleration  is  downward  (not  the  velocity  necessarily) 
we  make  p  negative  in  (5).     If  the  vessel  falls  freely,  p  =  —  g 
and  .'.p  =pa,  in  all  parts  of  the  liquid. 
Query  :  Suppose  p  downward  and  >  g. 
CASE  II.    Uniform  Rotation  about  a  Vertical  Axis. — If  the 
narrow  vessel  in  Fig.  473,  open  at  top  and  containing  a  liquid, 

be  kept  rotating  at  a  uniform  angu- 
lar velocity  GO  (see  §  110)  about  a 
vertical  axis  Z,  the  liquid  after  some 
oscillations  will  be  brought  (by  fric- 
tion) to  relative  equilibrium  (rotat- 
ing about  Z,  as  if  rigid).  Required 
the  form  of  the  free  surface  (evi- 
dently a  surface  of  revolution)  at 
each  point  of  which  we  know 

P=Pa- 

Let  0  be  the  intersection  of  the 
axis  Zwith  the  surface,  and  n  any  point  in  the  surface  ;  1)  being 


Fio.  473. 


UNIFORM   ROTATION    OF   LIQUID   IN   VESSEL.          543 

a  point  vertically  under  n  and  in  same  horizontal  plane  as  0. 
Every  point  of  the  small  right  prism  nb  (of  altitude  —  z  and 
sectional  area  dF)  is  describing  a  horizontal  circle  about  2,  and 
has  therefore  no  vertiacl  acceleration.  Hence  for  this  prism, 
free,  we  have  2Z  =  0;  i.e., 


dF.  pb  -  dF.pa  -  zdF.  y  =  0  .....    (1) 

Now  the  horizontal  right  prism  Ob  (call  the  direction  0  ...  5, 
X)  is  rotating  uniformly  about  a  vertical  axis  through  one  ex- 
tremity, as  if  it  were  a  rigid  body.  Hence  the  forces  acting 
on  it  must  be  equivalent  to  a  single  horizontal  force,  —  (*?Mp, 
(§1220*)  coinciding  in  direction  with  X.  [M=  mass  of  prism 
=  its  weight  -=-  <?,  and  p  =  distance  of  its  centre  of  gravity 
from  0  ;  here  p  =  \x  =  %  length  of  prism].  Hence  the 

xdF 
of  the  forces  acting  on  the  prism  Ob  must  =  —  caa 


tS 

But  the  forces  acting  on  the  two  ends  of  this  prism  are  their 
own  JT  components,  while  the  lateral  pressures  and  the  weights 
of  its  particles  have  no  X  compons.  ; 


JT?  JTT  .  /nx 

.:dF.pa-dF.pb  =  -  -  -  L.      .    .    (2) 


From  (1)  and  (2)  we  have 


where  v  =  cox  =  linear  velocity  of  the  point  n  in  its  circular 
path. 

[As  in  Case  I,  we  may  obtain  the  same  result  by  considering 
a  single  surface-particle  free,  and  would  derive  for  the  resultant 
force  acting  upon  it  the  value  dG  tan  a  in  a  horizontal  direc- 
tion and  intersecting  the  axis  of  rotation.  But  here  a  is  dif- 
ferent for  particles  at  different  distances  from  the  axis,  tan  a 

dz 
being  the  —  of  the  curve  On.     As  the  particle  is  moving  uni- 

formly in  a  circle  the  resultant  force  must  point  toward  the 


544  MECHANICS   OF   ENGINEERING. 

centre  of  the  circle,  i.e.,  horizontally,  and  have  a  value -, 

g    x 

where  x  is  the  radius  of  the  circle  [§  74,  eq.  (5)]  ; 


7  /~i    ,  w  \ji    \  uz/«*/ 1  CtZ  GO 

/.  dCr  tan  a  = - — J :  or  tan  a,  =  -- ,  =  — ; 

^       x  dx        g' 

fo;  or,s  =  -.|2.   .   Q.E.D. 

Hence  any  vertical  section  of  the  free  surface  through  the 
axis  of  rotation  Z  is  a  parabola,  with  its  axis  vertical  and  vertex 
at  0 ;  i.e.,  the  free  surface  is  a  paraboloid  of  revolution,  with 
Z  as  its  axis.  Since  cox  is  the  linear  velocity  v  of  the  point 
b  in  its  circular  path,  z  =  "height  due  to  velocity"  v  [§  52], 

EXAMPLE. — If  the  vessel  in  Fig.  473  makes  100  revol.  per 
minute,  required  the  ordinate  z  at  a  horizontal  distance  of  x  = 
4  inches  from  the  axis  (ft.-lb.-sec.  system).  The  angular  veloc- 
ity co  =  [2?r  100  -f-  60]  radians  per  sec.  [1ST.  B. — A  radian  = 
the  angular  space  of  which  3.1415926  .  .  .  make  a  half-revol., 
or  angle  of  180°].  With  x  =  -J  ft.  and  g  —  32.2, 


'•-2i   ---r3J^A=  ft.  =  2 

and  the  pressure  at  b  (Fig.  471)  is  (now  use  inch,  lb.,  sec.) 

f»9  K 

J  X  -        -  14.781  Ibs.  per  sq.  in. 


Prof.  Mendelejeff  of  Russia  has  recently  utilized  the  fact  an- 
nounced as  the  result  of  this  problem,  for  forming  perfectly 
true  paraboloidal  surfaces  of  plaster  of  Paris,  to  receive  by 
galvanic  process  a  deposit  of  metal,  and  thus  produce  specula 
of  exact  figure  for  reflecting  telescopes.  The  vessel  contain- 
ing the  liquid  plaster  is  kept  rotating  about  a  vertical  axis 
at  the  proper  uniform  speed,  and  the  plaster  assumes  the  de- 
sired shape  before  solidifying.  A  fusible  alloy,  melted,  may 
also  be  placed  in  the  vessel,  instead  of  liquid  plaster. 


RELATIVE   EQUILIBRIUM. 


545 


REMARK. — If  the  vessel  is  quite  full  and  closed  on  top,  ex- 
except  at  0'  where  it  communicates 
by  a  stationary  pipe  with  a  reser- 
voir, Fig.  474,  the  free  surface 
cannot  be  formed,  but  the  pres- 
sure at  any  point  in  the  water  is 
just  the  same  during  uniform  rota- 
tion, as  if  a  free  surface  were  formed 
with  vertex  at  0 ; 

See  figure  for  A0  and  2.  (In  subse- 
quent paragraphs  of  this  chapter 
the  liquid  will  be  at  rest.) 

FIG.  474. 

428a.  Pressure  on  the  Bottom  of  a  Vessel  containing  Liquid  at 
Best. — If  the  bottom  of  the  vessel  is  plane  and  horizontal,  the 
intensity  of  pressure  upon  it  is  the  same  at  all  points,  being 


FIG.  475. 


FIG.  476. 


p=pa-{-hy  (Figs.  475  and  476),  and  the  pressures  on  the  ele- 
ments of  the  surface  form  a  set  of  parallel  (vertical)  forces. 
This  is  true  even  if  the  side  of  the  vessel  overhangs,  Fig.  476, 
the  resultant  fluid  pressure  on  the  bottom  in  both  cases  being 


=  Fp-Fpa  =  Fhy. 


(1) 


(Atmospheric  pressure  is  supposed  to  act  under  the  bottom.) 
It  is  further  evident  that  if  the  bottom  is  a  rigid  homogeneous 
plate  and  has  no  support  at  its  edges,  it  may  be  supported  at  a 


546  MECHANICS   OF   ENGINEERING. 

single  point  (Fig.  477),  which  in  this  case  (horizontal  plate) 
is  its  centre  of  gravity.  This  point  is  called 
the  Centre  of  Pressure,  or  the  point  of  appli- 
cation of  the  resultant  of  all  the  fluid  pressures 
acting  on  the  plate.  The  present  case  is  such 
that  these  pressures  reduce  to  a  single  result- 
ant, but  this  is  not  always  practicable. 

EXAMPLE. — In  Fig.  476  (cylindrical  vessel 
FIG. 477.  containing    water),   given    h  =  20   ft.,   h^  = 

15  ft.,  rl  =  2  ft.,  ?\  =  4  ft.,  required  the  pressure  on  the  bot- 
tom, the  vertical  tension  in  the  cylindrical  wall  CA,  and  the 
hoop  tension  (§  426)  at  C.  (Ft.,  lb.,  sec.)  Press,  on  bottom  = 
Fhy  =  nrfhy  =  ?rl6  X  20  X  62.5  =  62857  Ibs. ;  while  the 
upward  pull  on  CA  = 

(TTT*  -  Ttr^h.y  =  ar(16  -  4)15  X  62.5  =  35357  Ibs. 

If  the  vertical  wall  is  t  =  -fa  inch  thick  at  C  this  tension  will 
be  borne  by  a  ring-shaped  cross-section  of  area  =  %nr£  (nearly) 
=  2;r48  X  TV  =  30.17  sq.  inches,  giving  (35357  -f-  30.17)  = 
about  1200  Ibs.  per  sq.  inch  tensile  stress  (vertical). 
The  hoop  tension  at  C  is  horizontal  and  is 

p"  =  rt(p  —pa)  -=-  t  (see  §  426),  where  p  —pa  X  h,y  ; 

„      48  X  15  X  12  X  (62.5  +  1728) 
S.p"  =  —  _ -- :  2  —  3125  Ibs.  per  scyin, 

TO"  ^^* 

(using  the  inch  and  pound). 

429.  Centre  of  Pressure. — In  subsequent  work  in  this  chapter, 
since  the  atmosphere  has  access  both  to  the  free  surface  of 
liquid  and  to  the  outside  of  the  vessel  walls,  and  pa  would  can- 
cel out  in  finding  the  resultant  fluid  pressure  on  any  elemen- 
tary area  dF  of  those  walls,  we  shall  write  : 

The  resultant  fluid  pressure  on  any  dF  of  the  vessel  wall  is 
normal  to  its  surface  and  is  dP  =  pdF=  zydF,  in  which  z 
is  the  vertical  distance  of  the  element  below  the  free  surface 
of  the  liquid  (i.e.,  z  =  the  "  head  of  water").  If  the  surface 
pressed  on  is  plane,  these  elementary  pressures  form  a  system 
of  parallel  forces,  and  may  be  replaced  by  a  single  resultant 


CENTRE   OF   PRESSURE. 


547 


(if  the  plate  is  rigid)  which  will  equal  their  sum,  and  whose 
point  of  application,  called  the  Centre  of  Pressure,  may  be 
located  by  the  equations  of  §  22,  put  into  calculus  form. 

If  the  surface  is  curved  the  elementary  pressures  form  a  sys 
tern  of  forces  in  space,  and  hence  (§  38)  cannot  in  general  be 
reduced  to  a  single  resultant,  but  to  two,  the  point  of  applica- 
tion of  one  of  which  is  arbitrary  (viz.,  the  arbitrary  origin, 
§38). 

Of  course,  the  object  of  replacing  a  set  of  fluid  pressures  by 
a  single  resultant  is  for  convenience  in  examining  the  equi- 
librium, or  stability,  of  a  rigid  body  the  forces  acting  on  which 
include  these  fluid  pressures.  As  to  their  effect  in  distorting 
the  rigid  body,  the  fluid  pressures  must  be  considered  in  their 
true  positions  (see  example  in  §  264),  and  cannot  be  replaced 
by  a  resultant. 

430.  Resultant  Liquid  Pressure  on  a  Plane  Surface  forming1 
Part  of  a  Vessel  Wall.  Co-ordinates  of  the  Centre  of  Pressure. — 
Fig.  478.  Let  AB  be  a  portion  (of  any  shape)  of  a  plane 
surface  at  any  angle  with  the 
horizontal,  sustaining  liquid 
pressure.  Prolong  the  plane 
of  AB  till  it  intersects  the  free 
surface  of  the  liquid.  Take 
this  intersection  as  an  axis  Y, 
O  being  any  point  on  Y.  The 
axis  X,  "1  to  Yj  lies  in  the 
given  plane.  Let  a  =  angle 
between  the  plane  and  the  free 
surface.  Then  x  and  y  are  the 
co-ordinates  of  any  elementary  FIG.  478. 

area  dF  of  the  surface,  referred  to  Xand  Y.  z  =  the  "  head 
of  water,"  below  the  free  surface,  of  any  dF.  The  pressures 
are  parallel. 

The  normal  pressure  on  any  dF  =  zydF\  hence  the  sum  of 
these,  =  their  resultant, 


(1) 


548  MECHANICS   OF   ENGINEERING. 

in  which  2  =  the  u  mean  2,"  i.e.,  the  z  of  the  centre  of  gravity 
G  of  the  plane  figure  AB,  and  F=  total  area  of  AB  \F  z  — 
fzdF,  from  eq.  (4),  §  23].  y  =  heaviness  of  liquid  (see  §  409). 
That  is,  the  total  liquid  pressure  on  a  plane  figure  is  equal 
to  tlie  weight  of  an  imaginary  prism  of  the  liquid  having  a 
base  =  area  of  the  given  figure  and  an  altitude  =  vertical 
depth  of  the  centre  of  gravity  of  the  figure  below  the  surface 
of  the  liquid.  For  example,  if  the  figure  is  a  rectangle  with 
one  base  (length  —  &)  in  the  surface,  and  lying  in  a  vertical 
plane, 

P  =  lh.  \hy  = 


Evidently,  if  the  altitude  be  increased,  P  varies  as  its  square. 

From  (1)  it  is  evident  that  the  total  pressure  does  not  de- 
pend on  the  horizontal  extent  of  the  water  in  the  reservoir. 

Now  let  xc  and  yc  denote  the  co-ordinates,  in  plane  YOX, 
of  the  centre  of  pressure,  C,  or  point  of  application  of  the  re- 
sultant  pressure  P,  and  apply  the  principle  that  the  sum  of 
the  moments  of  each  of  several  parallel  forces,  about  an  axis  1 
to  them,  is  equal  to  the  moment  of  their  resultant  about  the 
same  axis  [§  22].  First  taking  OY  as  an  axis  of  moments, 
and  then  OX,  we  have 

Pxc=  f*(zydF}x,    and  Pyc-=  f*(zydF}y.  .    (2) 

But  P  =  Fzy  —  Fx(§v&  a)y,  and  the  z  of  any  dF=  %  sin  a. 
Hence  eqs.  (2)  become  (after  cancelling  the  constant,  y  sin  a) 


__T  _ 

--  —  —  -  _  —  ,  ana  yc  — 


.     .     (6  ) 
Fx 

in  which  IY  =  the  "  mom.  of  inertia"  of  the  plane  figure  re- 
ferred to  Y  (see  §  85).  [K  B.  —  The  centre  of  pressure  as 
thus  found  is  identical  with  the  centre  of  oscillation  (§  117) 
and  the  centre  of  percussion  [§  113]  of  a  thin  homogeneous 
plate,  referred  to  axes  JTand  Y,  Y  being  the  axie  of  suspen- 
sion.] 

Evidently,  if  the  plane  figure  is  vertical  a  =  90°,  x  —  z  for 


CENTRE   OF   PRESSURE. 


549 


all  dF's,  and  x  =  z.     It  is  also  noteworthy  that  the  position 
of  the  centre  of  pressure  is  independent  of  a. 

NOTE. — Since  the  pressures  on  the  equal  dF's  lying  in  any 
horizontal  strip  of  the  plane  figure  form  a  set  of  equal  parallel 
forces  equally  spaced  along  the  strip,  and  are  therefore  equiva- 
lent to  their  sum  applied  in  the  middle  of  the  strip,  it  follows 
that  for  rectangles  and  triangles  with  horizontal  bases,  the 
centre  of  pressure  must  lie  on  the  straight  line  on  which  the 
middles  of  all  horizontal  strips  are  situated. 

431.  Centre  of  Pressure  of  Rectangles  and  Triangles  with  Bases 
Horizontal. — Since  all  the  dF's  of  one  horizontal  strip  have 
the  same  x,  we  may  take  the  area  of  the  strip  K 

for  dF  in  the  summation  fx*dF.     Hence  for 
the  rectangle  AB,  Fig  479,  we  have  from  eq. 


-KC=      b^dx      ^2.  A'- A*. 


0 

X      ifi 

1       -1 

Y 

j  

dx 

•  G 

c. 

W 

FIG.  479. 

~"» • ' — 

while  (see  note,  §  430)  yc  = 

When  the  upper  base  lies  in  the  surface,  A,  =  0,  and  xc  = 
JA2  —  -f-  of  the  altitude. 

For  a  triangle  with  its  base  horizontal  and  vertex  up,  Fig. 
480,  the  length  u  of  a  horizontal  strip  is  variable  and  dF= 

udx.     From  similar  triangles  u  =  — —(x  —  7^) ;  therefore 

An     A, 


550 


MECHANICS    OF   ENGINEERING. 


Also,  since  the  centre  of  pressure  must  lie  on  the  line  AB  join- 
ing the  vertex  to  the  middle  of  base  (see  note,  §  430),  we  easily 
determine  its  position. 

Evidently  for  hl  =  0,  i.e.,  when  the  vertex  is  in  the  surface, 
Jj> r /     xc  =  |Aa.     Similarly,  for   a   triangle  with 


_v*i  base  horizontal  and  vertex  down,  Fig.  481, 
we  find  that 


(3) 


If  the  base  is  in  the  surface,  A1  =  0  and 
FIG.  48i.  (3)  reduces  to  xc  =  |7i2. 

It  is  to  be  noticed  that  in  the  case  of  the  triangle  the  value 
of  xc  is  the  same  whatever  be  its  shape,  so  long  as  hl  and  A2 
remain  unchanged  and  the  base  is  horizontal.  If  the  base  is 
not  horizontal,  we  may  easily,  by  one  horizontal  line,  divide 
the  triangle  into  two  triangles  whose  bases  are  horizontal  and 
whose  combined  areas  make  up  the  area  of  the  first.  The  re- 
sultant pressure  on  each  of  the  component  triangles  is  easily 
found  by  the  foregoing  principles,  as  also  its  point  of  applica- 
tion. The  resultant  of  the  two  parallel  forces  so  determined 
will  act  at  some  point  on  the  line  joining  the  centres  of  pres- 
sures of  the  component  triangles,  this  point  being  easily  found 
by  the  method  of  moments,  while  the  amount  of  this  final  re- 
sultant pressure  is  the  sum  of  its  two  components,  since  the 
latter  are  parallel.  An  instance  of  this  procedure  will  be 
given  in  Example  3  of  §  433.  Similarly,  the  rectangle  of  Fig. 
479  may  be  distorted  into  an  oblique  parallelogram  with  hori- 
zontal bases  without  affecting  the  value  of  xc ,  nor  the  amount 
of  resultant  pressure,  so  long  as  hl  and  A2  remain  unchanged. 

432.  Centre  of  Pressure  of  Circle. — Fig.  482.  It  will  lie  on 
the  vertical  diameter.  Let  r  =  radius.  From  eq.  (3), 


Fx 


F 


(See  eq.  (4),  §  88,  and  also  §  91.) 


F:o.  482. 


x 


CENTRE   OF   PRESSURE. 


551 


FIG.  483. 


433.  Examples. — It  will  be  noticed  that  although  the  total 
pressure  on  the  plane  figure  depends  for  its  value  upon  the 
head,  a,  of  the  centre  of  gravity,  its  point  of  application  is  al- 
ways lower  than  the  centre  of  gravity. 

EXAMPLE  1. — If  6  ft.  of  a  vertical  sluice-gate,  4  ft.  wide, 
Fig.  483,  is  below  the  water-surface,  the  total 
water  pressure  against  it  is  (ft,  lb.,  sec. ;  eq. 
(1),§430) 

P  =  Fzy  =  6  X  4  X  3  X  62.5  =  4500  Ibs., 

and  (so  far  as  the  pressures  on  the  vertical 
posts  on  which  the  gate  slides  are  concerned) 
is  equivalent  to  a  single  horizontal  force  of 
that  value  applied  at  a  distance  xc  =  f  of 
6  =  4  ft.  below  the  surface  (§  431). 

EXAMPLE  2.— To  (begin  to)  lift  the  gate  in  Fig.  483,  the 
gate  itself  weighing  200  Ibs.,  and  the  coefficient  of  friction 
between  the  gate  and  posts  being/"  =  0.40  (abstract  numb.)  (see 
§  156),  we  must  employ  an  upward  vertical  force  at  least 

=  P'  =  200  +  0.40  X  4500  =  2000  Ibs. 

EXAMPLE  3. — It  is   required   to  find  the  resultant  hydro- 
static pressure  on  the  trapezoid  in  Fig.  483&  with  the  dimen- 
sions there  given  and  its  bases  horizontal ;  also  its  point  of  ap- 
plication, i.e.,  the  centre  of  pressure  of 

r~; the  plane  figure  in  the  position  there 

A      B  c       D  shown.    From  symmetry  the  C.  of  P.  will 

be  in  the  middle  vertical  of  the  figure, 
as  also  that  of  the  rectangle  B  CFE,  and 
that  of  the  two  triangles  ABE  and 
CDF  taken  together  (conceived  to  be 
shifted  horizontally  so  that  CF  and 
BE  coincide  on  the  middle  vertical, 
thus  forming  a  single  triangle  of  5  ft.  base,  and  having  the 
same  total  pressure  and  C.  of  P.  as  the  two  actual  triangles 
taken  together).  Let  Pl  =  the  total  pressure,  and  a?/  refer  to 
the  C,  of  P.,  for  the  rectangle ;  P9  and  a?/,  for  the  5  ft.  tri- 


EF-    5' 


FIG.  4S3a. 


552 


MECHANICS    OF   ENGINEERING. 


angle;  ht  =  4  ft.  and  A2  =  10  ft.  being  the  same  for  both. 
Then  from  eq.  (1),  §  430,  we  have  (with  the  ft.,  lb.,  and  sec.) 


Pl  =  30  X  1y  = 


and 


f  =  i  X  6  X  5  X  6y  =  90r; 


while  from  eqs.  (1)  and  (3)  of  §  431  we  have  also  (respectively) 

7.438  feet; 


2  1000  —  64 

3  °    100  -  16 


2     936 

3'~84~ 


-V,     9       

*•   ~  2" 


80       100 


228 


8+10 


2X18 


=  6.333  feet. 


The  total  pressure  on  the  trapezoid,  being  the  resultant  of 
Pl  and  P2 ,  has  an  amount  =  P1  +  P2  (since  they  are  parallel), 
and  has  a  lever-arm  scc  about  the  axis  OY  to  be  found  by  the 
principle  of  moments,  as  follows : 

P  V  +  />/  _  (210  X  7.438  +  90  X  6.33)x 
\  +  P,  ~(210  +  "^--  =  T 


*.  = 


The  total  hydrostatic  pressure  on  the  trapezoid  is  (for  fresh 
water) 

P  =  P,  +  P2  =  [210  +  90]  62.5  =  18T50  Ibs. 

EXAMPLE  4. — Required  the  horizontal  force  P',  Fig.  484,  to 
be  applied  at  N  (with  a  leverage  of  a'  =  30  inches  about  the 

fulcrum  M)  necessary  to  (begin 
to)  lift  the  circular  disk  AB  of 
radius  r  =  10  in.,  covering  an 
opening  of  equal  size.  NMAB 
is  a  single  rigid  lever  weighing 
#/  =  210  Ibs.  The  centre  of 
gravity,  G,  of  disk,  being  a  ver- 
tical distance  ~z  =  O'G  —  40 
inches  from  the  surface,  is  50 
inches  (viz.,  the  sum  of  OM  = 
k  =  20"  and  MG  =  30")  from 
axis  0  Y  ;  i.e.,  x  =  50  inches. 
The  centre  of  gravity  of  the 
whole  lever  is  a  horizontal  distance  J7,  =  12  inches,  from  M. 


FIG.  484. 


EXAMPLES  —  CENTRE   OF   PRESSURE.  553 

For  impending  lifting  we  must  have,  for  equilibrium  of  the 
lever, 

-k);      .    .     .     .    (1) 


where  P  =  total  water  pressure    on  circular  disk,   and  xc  = 
OC.     From  eq.  (1),  §  430,  (using  inch,  lb.,  and  sec.,) 


P  =  fzy  =  7tr*zy  =  arlOO  X  40  X  -ss"  =  454.6  Ibs. 

1728 


From  §432,  xc  =  OC  =  x  +  ~  =  50  +  1  .         =  50.5  in. 

4  2;  4: 


=  ~  [210  X  12  +  454.6  x  30.5]  =  546  Ibs. 
30 

434.  Example  of  Flood-gate. — Fig.  485.     Supposing  the  rigid 
double  gate  AD,  8  ft.  in  total  width,  to 
have  four  hinges ;  two  at  0,  and  two  at/ 
1  ft.  from  top  and  bottom  of  water  chan-    "^=j-'=i-='1  , 
nel ;  required  the  pressures  upon  them,        FT 
taking  dimensions  from  the  figure  (ft.,     ££  \ 
lb.,  sec.). 


Wat.  press.  =  P  —  Fzy 

=  72  X  4J-  X  62.5  =  20250 

pounds,  and  its  point  of  application  (cent,  of  press.)  is  a  dis- 
tance xc  =  f  of  9'  —  6'  from  O  (§  431).  Considering  the 
whole  gate  free  and  taking  moments  about  0,  we  shall  have 

(press,  at  f)xT  =  20250  x  5 ;  .-.  press,  at  /  =  14464  Ibs. 
(half  on  each  hinge  at/),  and 

/.  press,  at  e  =  P  —  press,  at/  =  5875  Ibs. 
(half  coming  on  eacli  hinge). 


554 


MECHANICS    OF    ENGINEERING. 


If  the  two  gates  do  not  form  a  single  rigid  body,  and  hence 
are  not  in  the  same  plane  when  closed,  a  wedge-like  or  toggle- 
joint  action  is  induced,  producing  much  greater  thrusts  against 
the  hinges,  and  each  of  these  thrusts  is  not  ~|  to  the  plane  of 
the  corresponding  gate.  Such  a  case  forms  a  good  exercise 
for  the  student. 

435.  Stability  of  a  Vertical  Rectangular  Wall  against  Water 
Pressure  on  One  Side. — Fig.  486.  All  dimensions  are  shown  in 

the  figure,  except  I,  which  is  the  length 
of  wall  ~]  to  paper.    Supposing  the  wall 
to  be  a  single  rigid  block,  its  weight  G' 
—  Vh'ly'  (y'  being  its  heaviness  (§  7), 
and  I  its  length).     Given  the  water 
depth  —  A,  required  the  proper  width 
bf  for  stability.     For  proper  security  : 
First,  the  resultant  of  G'  and  the 
Fl°-  4g6.  water-pressure  P  must  fall  within  the 

base  BD  (or,  which  amounts  to  the  same  thing),  the  moment 
of  G  about  D,  the  outer  toe  of  the  wall,  must  be  numerically 
greater  than  that  of  P ;  and 

Secondly,  P  must  be  less  than  the  sliding  friction/Vjr'  (see 
§  156)  on  the  base  BD. 

Thirdly,  the  maximum  pressure  per  unit  of  area  on  the 
base  must  not  exceed  a  safe  value  (compare  §  348). 

NowJ3  =  Fzy  =  Til  —  y  =  —  h*ly  (y  —  heaviness  of  water) ; 

2  2 

and  xc  —  -f  A. 

Hence  for  stability  against  tipping  about  D, 

P^h  must  oe  <  G'^b' ;    i.e.,  \tfly  <  ^b'*h'ly' ;     .     (1) 
while,  as  to  sliding  on  the  base, 

i.e.,  $h*ly  <fb'h'lyf.     .     .     (2) 


As  for  values  of  the  coefficient  of  friction,/",  on  the  base  of 
wall,  Mr.  Fanning  quotes  the  following  among  others,  from 
various  authorities : 


STABILITY   OF   WALL. 


555 


u  a 

a  u 

a  u 

a  a 


For  point-dressed  granite  on  dry  clay,  f  —  0.51 

"            "       "    moist  clay,  0.33 

"             "        "   gravel,  0.58 

"            "       "    smooth  concrete,  0.62 

"             "        u    similar  granite,  O.TO 

.    For  dressed  hard  limestone  on  like  limestone,  0.38 

"           4>             "         "    brickwork,  0.60 

For  common  bricks  on  common  bricks,  0.64 

To  satisfactorily  investigate  the  third  condition  requires  the 
detail  of  the  next  paragraph. 

436,  Parallelopipedical  Reservoir  Walls.  More  Detailed  and 
Exact  Solution.  —  If  (1)  in  the  last  paragraph  were  an  exact 
equality,  instead  of  an  inequality, 
the  resultant  R  of  P  and  Gf 
would  pass  through  the  corner 
J9,  tipping  would  be  impending, 
and  the  pressure  per  unit  area  at 
D  would  be  theoretically  infinite. 
To  avoid  this  we  wish  the  wall 
to  be  wide  enough  that  the  re- 
sultant 7?,  Fig.  487,  may  cut 
BD  in  such  a  point,  E'  ,  as  to  cause  the  pressure  per  unit  area, 
pm,  at  D  to  have  a  definite  safe  value  (for  the  pressure  pm  at 
D,  or  quite  near  _Z>,  will  evidently  be  greater  than  elsewhere 
on  BD  ;  i.e.,  it  is  the  maximum  pressure  to  be  found  on  BD). 
This  may  be  done  by  the  principles  of  §§  346  and  362. 

First,  assume  that  R  cuts  BD  outside  of  the  'middle  third; 
i.e.,  that 

VE!,  =  nb',  >  #'(<"•  n  >  |)  ; 


FIG.  487. 


where  n  denotes  the  ratio  of  the  distance  of  E'  from  the  mid- 
dle of  the  base  to  the  whole  width,  £>',  of  base.  Then  the  pres- 
sure (per  unit  area)  on  small  equal  elements  of  the  base  BD 
(see  §  346)  may  be  considered  to  vary  as  the  ordinates  of  a 
triangle  MND  (the  vertex  M  being  within  the  distance  BD), 
and  E~D  will  =  \MD  ;  i.e., 


556  MECHANICS   OF  ENGINEERING. 


The  mean  pressure  per  unit  area,  on 


and  hence  the  maximum  pressure  (viz.,  at  D\  being  double 
the  mean,  is 

pm  =  iff'  -=-  [8J7(i  -  n)] ;      .     .     .     .     (0) 

and  if  pmis  to  equal  ^'(see  §§  201  and  203),  a  safe  value  for  the 
crushing  resistance,  per  unit  area,  of  the  material,  we  shall 
have 

&7(i  -n)Cf  =  \G'  =  ty'k'ly', 

i     2  hy 

•'•  Vs  it  fan (1) 

To  find  &',  knowing  w,,  we  put  the  ^(moms.)  of  the  Gr  and  P 
at  E,  about  E',=-  zero  (for  the  only  other  forces  acting  on 
the  wall  are  the  pressures  of  the  foundation  against  it,  along 
MD ;  and  since  the  resultant  of  these  latter  passes  through  E\ 
the  sum  of  their  moments  about  E'  is  already  zero) ;  i.e., 

QW  -  pk  =  o or    nbnh'l= 


Having  obtained  5r,  we  must  also  ascertain  if  P  is  <fG-',  the 
friction  ;  i.e.,  if  P  is  <  fb'h'ly'.  If  not,  &'  must  be  still  further 
increased.  (Or,  graphically,  the  resultant  of  Gf  and  P  must 
not  make  an  angle  >  0,  the  angle  of  friction,  with  the  ver- 
tical. 

If  n,  computed  from  (1),  should  prove  to  be  <  £,  our  first 
assumption  is  wrong,  and  we  therefore  assume  n  <  ^,  and  pro- 
ceed thus : 

Secondly,  n  being  <  £  (see  §§  346  and   362),  we  have  a 


STABILITY   OF   KESfcRVOIK   WALLS.  557 

trapezoid  of  pressures,  instead  of  a  triangle,  on  BD.  Let  the 
pressure  per  unit  area  at  D  be  pm  (the  maximum  on  base). 
The  whole  base  now  receives  pressure,  the  mean  pressure  (per 
unit  area)  being  =  G-'  -5-  [5'Z]  ;  and  therefore,  from  §  362, 
Case  I,  we  have 

;    .....    (Oa) 


and  since,  here,  G'  =  Vh'ly'  ',  we  may  write 

pm  =  (Qn  +  1)A  y. 
For  safety  as  to  crushing  resistance  we  put 

6(n  +  l)hy  =  C';  whence  n  =  |  ["^-7  -  ll  .    .     (la) 

Having  found  n  from  eq.  (la),  we  determine  the  proper 
width  of  base  I'  from  eq.  (2),  in  case  the  assumption  n  <  -J-  is 
verified. 

EXAMPLE.—  In  Fig.  486,  let  hf  =  12  ft.,  h  =  10  ft.,  while 
the  masonry  weighs  (y1  =)  150  Ibs.  per  cub.  ft.  Supposing 
it  desirable  to  bring  no  greater  compressive  stress  than  100  Ibs. 
per  sq.  inch  (=  14400  Ibs.  per  sq.  ft.)  on  the  cement  of  the 
joints,  we  put  C'  14400,  using  the  ft.-lb.-sec.  system  of  units. 

Assuming  n  >  -J-,  we  use  eq.  (1),  and  obtain 

_  1  _  2    12  X  150  _    _5_ 
~  2  "  3  '     14:400  "  ~  12' 

which  is  >  -J-  ;  hence  the  assumption  is  confirmed,  also  the 
propriety  of  using  eq.  (1)  rather  than  (la). 
Passing  to  eq.  (2),  we  have 


But,  as  regards  frictional  stability,  we  find  that,  wiibf  =  0.30,, 
a  low  value,  and  V  =  3.7ft.  (ft.,  lb.,  sec.), 


558  MECHANICS    OF    ENGINEERING. 

P_         ±h*ly     _  _  100  X  62.5  _  =  15. 
f  ~fbWl?  ""  2X0.3  X  3.7  X  12  X  150  ~ 


which  is  greater  than  unity,  showing  the  friction  to  be  insuf- 
ficient to  prevent  sliding  (with  /=0.30);  a  greater  width 
must  therefore  be  chosen,  for  frictional  stability. 

If  we  make  n  =  -J-,  i.e.,  make  R  cut  the  base  at  the  outer 
edge  of  middle  third  (§  362),  we  have,  from  eq.  (2), 


/  =  iox     •   62'5X] 


X  12  X  150 

and  the  pressure  at  D  is  now  of  course  well  within  the  safe 
limit ;  while  as  regards  friction  we  find 

P  -r-fG'  =  0.92,  <  unity, 

and  therefore  the  wall  is  safe  in  this  respect  also. 

With  a  width  of  base  =  3.7  feet  first  obtained,  the  portion 
MD,  Fig.  487,  of  the  base  which  receives  pressure  [according 
to  Navier's  theory  (§  346)]  would  be  only  0.92  feet  in  length, 
or  about  one  sixth  of  the  base,  the  portion  BM  tending  to 
open,  and  perhaps  actually  suffering  tension,  if  capable  (i.e.,  if 
cemented  to  a  rock  foundation),  in  which  case  these  tensions 
should  properly  be  taken  into  account,  as  with  beams  (§  295), 
thus  modifying  the  results. 

It  has  been  considered  safe  by  some  designers  of  high 
masonry  dams,  to  neglect  these  possible  tensile  resistances,  as 
has  just  been  done  in  deriving  ~b'  •=.  3.7  feet ;  but  others,  in 
view  of  the  more  or  less  uncertain  and  speculative  character  of 
Navier's  theory,  when  applied  to  the  very  wide  bases  of  such 
structures,  prefer,  in  using  the  theory  (as  the  best  available), 
to  keep  the  resultant  pressure  within  the  middle  third  at  the 
base  (and  also  at  all  horizontal  beds  above  the  base),  and  thus 
avoid  the  chances  of  tensile  stresses. 

This  latter  plan  is  supported  by  Messrs.  Church  and  Fteley, 
as  engineers  of  the  proposed  Quaker  Bridge  Dam  in  connec- 
tion with  the  New  Croton  Aqueduct  of  New  York  City,  in 
iheir  report  of  1887.  See  §  438. 


RESEEVOIR   WALLS. 


437.  Wall  of  Trapezoidal  Profile.  Water-face  Vertical- 
Economy  of  material  is  favored  by  using  a  trapezoidal  profile,, 
Fig.  488.  With  this  form  the 
stability  may  be  investigated  in 
a  corresponding  manner.  The 
portion  of  wall  above  each 
horizontal  bed  should  be  ex- 
amined similarly.  The  weight 
G'  acts  through  the  centre  of 
gravity  of  the  whole  mass. 

Detail.— Let  Fig.  488  show 
the  vertical  cross-section  of  a 
trapezoidal  wall,  with  notation 
for  dimensions  as  indicated  ;  the 
portion  considered  having  a  length  =  I,  ~]  to  the  paper.  Let 
y  =  heaviness  of  water,  y'  that  of  the  masonry  (assumed  homo- 
geneous), with  n  as  in  §  436. 

Fora  triangle  of  pressure,  MD,  on  the  base,  i.e.,  withn  >  -J, 
or  resultant  falling  outside  the  middle  third  (neglecting  pos- 
sibility of  tensile  stresses  on  left  of  M\  if  the  intensity  of 
pressure  j9TO  at  D  is  to  —  C'  (§  203),  we  put,  as  in  §  436, 


FIG.  488. 


-  n\  C'  = 


.e.   = 


whence 


For  a  trapezoid  of  pressure,  i.e.  with  n  <  -J-,  or  the  resultant 
of  P  and  G'  falling  within  the  middle  third,  we  have,  as  be- 
fore (§  362,  Case  I), 


~ 


whence 


n  =  - 


i 
n  = 


560  MECHANICS   OF   ENGINEERING. 

From  the  geometry  of  the  figure,  having  joined  the  middles 
of  the  two  bases,  we  have 


{§  26,  Prob.  6),  and,  by  similar  triangles,  OT  :  KV  ::gO:hr, 
whence 


The  lines  of  action  of  £'  and  P  meet  at  E,  and  their  result- 
ant cuts  the  base  in  some  point  E'.  The  sum  of  their  moments 
about  E'  should  be  zero,  i.e.,  P  .  ±h  —  Gf  .  'OE'\  that  is,  (see 
eq.  (a)  above,  and  eq.  (1),  §  430,) 


i.e.,  cancelling, 

+  V')         (2)' 


Hence  we  have  two  equations  for  finding  two  unknowns 
viz.:  (l)'and  (2)'  when  n  >  -J-  ;  and  (la)7  and  (2)r  when  /i  <  -J-. 

For  dams  of  small  height  (less  than  40  ft.,  say),  if  we  im- 
mediately put  n  =  -J-,  thus  restricting  the  resultant  pressure  to 
the  edge  of  middle  third,  and  solve  (2)7  for  J7,  1)"  being  as- 
sumed of  some  proper  value  for  a  coping,  foot-  walk,  or  road- 
way, while  h'  may  be  taken  enough  greater  than  h  to  provide 
against  the  greatest  height  of  waves,  from  2.5  to  6  ft.,  the 
value  of  pm  at  D  will  probably  be  <  C'.  In  any  case,  for  a 
value  of  n  =,  or  <,  £  we  put  pm  for  <7'in  equation  (la)7  and 
solve  for  pm  ,  to  determine  if  it  is  no  greater  than  C'  . 

Mr.  Fanning  recommends  the  following  values  for  C'  (in  Us. 


KESEKVOIR  WALLS.  561 

per  sq.foof)  with  coursed  rubble  masonry  laid  in  strong  mor- 
tar: 

For  Limestone.      Sandstone.       Granite.  Brick. 

O       =         50,000         50,000         60,000         35,000 


of  ) 

in  > 

ft.     f 


152  132  154  120 

Ibs.  per  cub." 

As  tofrictional  resistance,  P  must  be  <fG'\  i.e., 


If  the  base  is  cemented  to  a  rock  foundation  with  good 
material  and  workmanship  throughout,  Messrs.  Church  and 
Fteley  (see  §  436)  consider  that  the  wall  may  be  treated  as 
amply  safe  against  sliding  on  the  base  (or  any  horizontal  bed), 
provided  the  other  two  conditions  of  safety  are  already  satis- 
fied. 

438,  Triangular  Wall  with  Vertical  Water-face.—  Making 
l>"  =  0  in  the  preceding  article,  the  trapezoid  becomes  a  right 
triangle,  and  the  equations  reduce  to  the  following  : 


and 

l-}forn<k     ...    (la/7 


(pm  not  to  exceed  C'  in  any  case)  ;  while  to  determine  the 
breadth  of  base,  b',  after  n  is  computed  [or  assumed,  for  small 
height  of  wall],  we  have  from  eq.  (2)r,  (for  n  <  £,) 


(2a)" 
Also,  for  frictional  stability, 

tftly  must  be  <  \fhlltlyf  .....     (3)" 


562  MECHANICS   OF  ENGINEERING. 

439.  High  Masonry  Dams.  —  Although  the  principle  of  the 
arch  may  be  utilized  for  vertical  stone  dikes  of  small  height 

(30  to  50  feet)  and  small  span,  for 
greater  heights  and  spans  the 
formula  for  hoop  tension,  §  426  (or 
rather,  here,  "  hoop  compression"), 
on  the  vertical  radial  joints  of  the 
horizontal  arch  rings,  Fig.  489,  calls 
for  so  great  a  radial  thickness  of 
joint  in  the  lower  courses,  that 
straight  dikes  (or  "gravity  dams") 
are  usually  built  instead,  even 
where  firm  rock  abutments  are  available  laterally. 

For  example,  at  a  depth  of  100  feet,  where  the  hydrostatic 
pressure  is  hy  =  100  X  62.5  =  6250  Ibs.  per  sq.  ft.,  if  we  as- 
sume for  the  voussoirs  a  (radial,  horizontal)  thickness  =  4  ft., 
with  a  (horizontal)  radius  of  curvature  r  =  100  feet,  we  shall 
find  a  compression  between  their  vertical  radial  faces  of  (ft., 
lb.,  sec.) 

100  X6250 


or  1085  Ibs.  per  sq.  inch  ;  far  too  great  for  safety,  even  if  there 
were  no  danger  of  collapse,  the  dike  being  short.  If  now  the 
thickness  is  increased,  in  order  to  distribute  the  pressure  over 
a  greater  surface,  we  are  met  by  the  fact  that  the  formula  for 
"  hoop  compression"  is  no  longer  strictly  applicable,  the  law  of 
distribution  of  pressure  becoming  very  uncertain;  and  even 
supposing  a  uniform  distribution  over  the  joint,  the  thickness 
demanded  for  proper  safety  against  crushing  is  greater  than 
for  a  straight  dam  ("  gravity  dam"}  at  a  very  moderate  depth 
below  the  water  surface,  unless  the  radius  of  curvature  of  arch 
can  be  made  small.  But  the  smaller  the  radius  the  more  does 
the  darn  encroach  on  the  storage  capacity  of  the  reservoir,  while 
in  no  case,  of  course,  can  it  be  made  smaller  than  half  the  span. 
Another  point  is,  that  as  masonry  is  not  destitute  of  elas- 
ticity, the  longer  the  span  the  more  unlikely  is  it  that  the 
parts  of  the  arch  will  "  close  up"  properly,  and  develop  the 


HIGH  MASONKY  DAMS.  563 

abutment  reactions  when  the  water  is  first  admitted  to  the 
reservoir ;  which  should  occur  if  it  is  to  act  as  an  arch  instead 
of  bj  gravity  resistance. 

For  these  reasons  the  engineers  of  the  proposed  Quaker 
Bridge  Dam  reported  unfavorably  to  the  plan  of  a  curved  de- 
sign for  that  structure,  and  recommended  that  a  straight  dam 
be  built.  See  reference  in  §  436.  According  to  their  designs 
this  dam  is  to  be  258  feet  in  height  (which  exceeds  by  about  90 
feet  the  height  of  any  dam  previously  built),  about  1400  feet 
in  length  at  the  top,  and  216  feet  in  width  at  the  lowest 
point  of  base,  joining  the  bed-rock. 

More  recently,  however  (1888),  a  board  of  experts,  specially 
appointed  for  the  purpose,  having  examined  a  number  of  dif- 
ferent plans,  have  reported  favorably  to  the  adoption  of  a 
jcurved  form  for  the  dam,  as  offering  greater  resistance  under 
extraordinary  circumstances  (impact  of  ice-floes,  earthquakes, 
etc.),  on  account  of  its  arched  form  (though  resisting  by 
gravity  action  under  usual  conditions)  than  a  straight  struc- 
ture ;  and  also  as  more  pleasing  in  appearance. 

Fig.  490  shows  the  profile  of  a  straight  high  masonry  darn 
as  designed  at  the  present  day.  Assuming  a  width  l>"  =  from 
6  to  22  feet  at  the  top,  and  a  sufficient  h"  (see  figure)  to  ex- 
ceed the  maximum  height  of  waves,  the  up-stream  outline 
A  CM  is  made  nearly  vertical  and  perhaps  somewhat  concave, 
while  the  down-stream  profile  BDN,  by  computation  or 
graphical  trial,  or  both,  is  so  formed  that  when  the  reservoir  is 
full  the  resultant  R,  of  the  weight 
G  of  the  portion  AECD  of  ma- 
sonry  above  each  horizontal  bed,  as  1  ^j 
(7Z>,  and  the  hydrostatic  pressure  P 
on  the  corresponding  up-stream  face 
AC,  shall  cut  the  bed  CD  in  such  a 
point  E'  as  not  to  cause  too  great 
compression  pm  at  the  outer  edge  D 
(not  over  85  Ibs.  per  sq.  inch  accord- 
ing to  M.  Krantz  in  "  Keservoir  FIG.  490. 
Walls"),  pm  being  computed  by  one  of  the  equations  [(0)  and 
(Oa)  of  1 436] 


5b*4  MECHANICS    OF    ENGINEERING. 

For  E'  outside  the  middle  third  )  26r 

\  <Y)          .    (]_)"" 

and  neglecting  tension  )    .•  3.CD.l(%)—n 

For  E'  inside  middle  third  1  . . pm  =  ^n  +  1^ ;     (la)'" 

>  CD. I 

where  I  =  length  of  wall  1  to  paper,  usually  taken  =  one  foot, 
or  one  inch,  according  to  the  unit  of  length  adopted ;  for  n, 
see  §  436. 

Nor,  when  the  reservoir  is  empty  and  the  water  pressure 
lacking,  must  the  weight  G  resting  on  each  bed,  as  CD,  cut 
the  bed  in  a  point  E"  so  near  the  edge  C  as  to  produce  exces- 
sive pressure  there  (computed  as  above).  The  figure  shows 
the  general  form  of  profile  resulting  from  these  conditions. 
The  masonry  should  be  of  such  a  character,  by  irregular  bond- 
ing in  every  direction,  as  to  make  the  wall  if  possible  a  mono- 
lith. For  more  detail  see  next  paragraph. 

440.  Quaker  Bridge  Dam  (on  the  New  Croton  Aqueduct). — 
Attempts,  by  strict  analysis,  to  determine  the  equation  of  the 
curve  BN,  AM  being  assumed  straight,  so  as  to  bring  the 
point  E'  at  the  outer  edge  of  the  middle  third  of  its  joint,  or 
to  make  the  pressure  at  D  constant  below  a  definite  joint,  have 
failed,  up  to  the  present  time;  but  approximate  and  tentative 
methods  are  in  use  which  serve  all  practical  purposes.  As  an 
illustration  the  method  set  forth  in  the  report  on  the  Quaker 
Bridge  Dam  will  be  briefly  outlined ;  this  method  confines  E' 
to  the  middle  third. 

The  width  AB  —  I"  is  taken  —  22'  for  a  roadway,  and  h"  = 
7  ft.  The  profile  is  made  a  vertical  rectangle  from  A  down 
to  a  depth  of  33  ft.  below  the  water  surface  (reservoir  full). 
Combining  the  weight  of  this  rectangle  of  masonry  with  the 
corresponding  water  pressure  (for  a  length  of  wall  =  one  foot), 
we  find  the  resultant  pressure  conies  a  little  within  the  outer 
edge  of  the  middle  third  of  the  base  of  the  rectangle,  while 
pm  is  of  course  small. 

The  rectangular  form  of  profile  might  be  continued  below 
this  horizontal  joint,  as  far  as  complying  with  the  middle 


QUAKER    BRIDGE    DAM.  565 

third  requirement,  and  the  limitation  of  pressure-intensity,  are 
concerned ;  but,  not  to  make  the  widening  of  the  joints  too 
abrupt  in  a  lower  position  where  it  would  be  absolutely  re- 
quired, a  beginning  is  made  at  the  joint  just  mentioned  by 
forming  a  trapezoid  between  it  and  a  joint  11  ft.  farther  down, 
making  the  lower  base  of  the  latter  of  some  trial  width,  which 
can  be  altered  when  the  results  to  which  it  gives  rise  become 
evident.  Having  computed  the  weight  of  this  trapezoid  and 
constructed  its  line  of  action  through  the  centre  of  gravity  of 
the  trapezoid,  the  value  of  the  resultant  G  of  this  weight  and 
that  of  the  rectangle  is  found  (by  principle  of  moments  or  by 
an  equilibrium  polygon)  in  amount  and  position,  and  combined 
with  the  water  pressure  of  the  corresponding  44  ft.  of  water  to 
form  the  force  _Z?,  whose  point  of  intersection  with  the  new 
joint  or  bed  (lower  base  of  trapezoid)  is  noted  and  the  value  of 
pm  computed.  These  should  both  be  somewhat  nearer  their 
limits  than  in  the  preceding  joint.  If  not,  a  different  width 
should  be  chosen,  and  changed  again,  if  necessary,  until  satis- 
factory. Similarly,  another  layer,  11  ft.  in  height  and  of 
trapezoidal  form,  is  added  below  and  treated  in  the  same  way ; 
and  so  on  until  in  the  joint  at  a  depth  of  66  ft.  from  the 
water  surface  a  width  is  found  where  the  point  E'  is  very 
close  upon  its  limiting  position,  while pm  is  quite  a  little  under 
the  limit  set  for  the  upper  joints  of  the  dam,  8  tons  per  square 
foot.  For  the  next  three  11  ft.  trapezoidal  layers  the  chief 
governing  element  is  the  middle-third  requirement,  E'  being 
kept  quite  close  to  the  limit,  while  the  increase  of  pm  to  7.95 
tons  per  sq.  ft.  is  unobjectionable;  also,  we  begin  to  move 
the  left-hand  edge  to  the  left  of  the  vertical,  so  that  when  the 
reservoir  is  empty  the  point  E"  shall  not  be  too  near  the  up- 
stream edge  C. 

Down  to  a  depth  of  about  200  ft.  the  value  of  pm  is  allowed 
to  increase  to  10.48  tons  per  sq.  ft.,  while  the  position  of  Er 
gradually  retreats  from  the  edge  of  its  limit.  Beyond  200  ft. 
depth,  to  prevent  a  rapid  increase  of  width  and  consequent 
extreme  flattening  of  the  down-stream  curve,  pm  is  allowed 
to  mount  rapidly  to  16.63  tons  per  sq.  ft.  (=231  Ibs.  per 
sq.  in.),  which  value  it  reaches  at  the  point  ^Tof  the  base  of 


566  MECHANICS    OF   ENGINEEEING. 

the  dam,  which  has  a  width  =  216  ft.,  and  is  258  feet  below 
the  water  surface  when  the  reservoir  is  full. 

The  heaviness  of  the  masonry  is  taken  as  yf  =  156.25  Ibs. 
per  cubic  foot,  just  f  of  y  =  62.5  Ibs.  per  cub.  foot,  the  heavi- 
ness taken  for  water.  *, 

When  the  reservoir  is  empty,  we  have  the  weight  G  of  the 
superincumbent  mass  resting  on  any  bed  CD,  and  applied 
through  the  point  E" ;  the  pressure  per  unit  area  at  C  can 
then  be  computed  by  eq.  (lfl)'",  §439,  n  being  the  quotient  of 
(faCD  —  CE")  -r-  Cl)  for  this  purpose.  In  the  present  case 
we  find  E"  to  be  within  middle  third  at  all  joints,  and  the 
pressures  at  C  to  be  under  the  limit. 

For  further  details  the  reader  is  referred  to  the  report  itself 
(reprinted  in  Engineering  News,  January,  1888,  p.  20).  The 
graphic  results  were  checked  by  computation,  Wegmann's 
method,  applied  to  each  trapezoid  in  turn. 

441.  Earthwork  Dam,  of  Trapezoidal  Section. — Fig.  491.    It  is 
.    &  ^  required  to  find  the  conditions  of  sta- 

bility of  the  straight  earthwork  dam 
ABDE,  whose  length  —I,  L  to 
paper,  as  regards  sliding  horizontally 
on  the  plane  AE\  i.e.,  its  frictional 
stability.  With  the  dimensions  of 
the  figure,  y  and  y'  being  the  heavi- 
FIG.  49i.  nesses  of  the  water  and  earth  respec- 

tively (see  §  7),  we  have 


Weight  of  dam  =  G,  =  vol.  X  /  =  lh$  +  ^(a,  +  c)]y'.  (1) 

Eesultant  water  press.  =  P  =  Fzy  =  OA  X  I  X  %hy.    .  (2) 
Horiz.  comp.  of  P  =  II  =  P  sin  a 

=  [OA  sin  a] \Uy  -  %tfly.      .*    .     .  (3) 


From  (3)  it  is  evident  that  the  horizontal  component  of  P  is 
just  the  same,  viz.,  =  hi .  %hy,  as  the  water  pressure  would  be 
on  a  vertical  rectangle  equal  to  the  vertical  projection  of  OA 


EAKTHWORK    DAM. 


567 


and  with  its  centre  of  gravity  at  the  same  depth  (JA).     Com- 
pare §416.     Also, 

Vert.  comp.  of  P  =  V  =  P  cos  a 

=  [OA  cos  a]%kly  =*%akly,        ...     (4) 


and  is  the  same  as  the  water  pressure  on  the  horizontal  projec- 
tion of  OA  if  placed  at  a  depth  =  O'G  =  %h. 

For  stability  against  sliding,  the  horizontal  component  of  P 
must  be  less  than  the  friction  due  to  the  total  vertical  pressure 
on  the  plane  AE,  viz.,  G1  +  F;  hence  if  /"is  the  coefficient  of 
friction  on  AE,  we  must  have  H  <f  \_Gl  -f-  "F],  i.e.  (see  above), 


However,  if  the  water  leak  under  the  dam  on  the  surface  AE, 
so  as  to  exert  an  upward  hydrostatic  pressure 


(to  make  an  extreme  supposition,)  the  friction  will  be  only 

=/[#,+  v-  v'i 

and  (5)  will  be  replaced  by 

//</[£,  +  F--F']  ......    (6) 

Experiment  shows  (Weisbach)  that  with  y=0.33  computa- 
tions made  from  (6)  (treated  as  a  bare  equality)  give  satisfactory 
results. 

EXAMPLE.—  (Ft.,  lb.,  sec.)  With  /  =  0.33,  h  =  20  ft.,  h,  = 
22  ft.,  a  =  24  ft.,  a,  —  26.4  ft.,  and  c  =  30  ft.,  we  have,  mak- 
ing (6)  an  equality,  with  y'  —  %y, 


.2)2+J  (24x20)  -(26.4  +  b+  30)20]; 
whence,  solving  for  J,  the  width  of  top,  b  =  10.3  feet. 


568 


MECHANICS    OF   ENGINEERING. 


M 


Oi 


442.  Liquid  Pressure  on  Both  Sides  of  a  Gate  or  Rigid  Plate. — 
The  sluice-gate  AB,  for  example,  Fig.  492,  receives  a  pressure, 

PI  ,  from  the  "  head-water"  M,  and 
an  opposing  pressure  Pa  from  the 

"tail-water"  N.     Since    these  two 

-^^rTr^z^y-  horizontal  forces  are  not  in  the  same 
;  line,  though  parallel,  their  resultant 
ft,  which  =  P1  —  P2 ,  acts  horizon- 
="  tally  in  the  same  plane,  but  at  a  dis- 
FIG.  493.  tance  below  (?,  =  u,  which  we  may 

find  by  placing  the  moment  of  It  about  Ol ,  equal  to  the  alge- 
braic sum  of  those  of  Pl  and  P9  about  0  . 


(1) 


(7,  and  (73  are  the  respective  centres  of  pressure  of  the  surfaces 
O^B  and  OJ3,  and  u  =  distance  of  E  from  01 ,  while  A  =  dif- 
ference of  level  between  head  and  tail  waters.  If  the  surfaces 
O^  and  OJE>  are  both  rectangular, 

EXAMPLE. — Let  the  dimensions  be  as  in  Fig.  493,  both  sur- 
A  faces    under    pressure  being  rect- 

..,. te.    angular  and  8  ft.  wide.     Then  (ft., 

or  (§  430) 


.,  sec.)  R  =  Pl  — 
=  [12X8X6  —  8X8X 
=  20000  Ibs.  =10  tons; 
while  from  ex.  (2) 


Q! 


FIG.  493. 


=  [12X8X6X8-8X8X  4(9i)]62.5 
20000 

That  is,  u  =  6.93  feet,  which  locates  C.     Hence  the  pressure 
of  the  gate  upon  its  hinges  or  other  support  is  the  same  (aside 


WATER   ON  BOTH   SIDES   OF   GATE. 


569 


from  its  own  weight),  provided  it  is  rigid,  as  if  the  single 
horizontal  force  R  —  10  tons  acted  at  the  point  (7,  2.93  ft.  be- 
low the  level  of  the  tail-water  surface. 

443.  If  the  plate,  or  gate,  is  entirely  below  the  tail-water 
surface,  the  resultant  pressure  is  applied  in  the  centre  of  gravity 
of  the  plate.  —  Proof  as  follows  :  Conceive  the  surface  to  be 
divided  into  a  great  number  of  small  equal  areas,  each  =  dF\ 
then,  the  head  of  water  of  any  dF  being  =  xt  on  the  head- 
water side,  and  =  a?2  on  the  tail-water  side,  the  resultant  pres- 
sure on  the  dF  is  ydF(xl  —  a?a)  =yhdJ^t  in  which  h  is  the 
difference  of  level  between  head  and  tail  water.  That  is,  the 
resultant  pressures  on  the  equal  dF's  are  equal,  and  hence 
form  a  system  of  equal  parallel  forces  distributed  over  the  plate 
in  the  same  manner  as  the  weights  of  the  corresponding  por- 
tions of  the  plate  ;  therefore  their  single  resultant  acts  through 
the  centre  of  gravity  of  the  plate  ;  Q.  E.  D.  This  single  re- 
sultant =fyhdF  =  yhfdF  =  Fhy. 

EXAMPLE.  —  Fig.  494.  The  resultant  pressure  on  a  circular 
disk  ab  of  radius  =  8  inches,  (in 
the  vertical  partition  OK,}  which 
has  its  centre  of  gravity  3  ft. 
below  the  tail-water  surface,  with 
h  =  2  ft,  is  (ft.,  lb.,  sec.) 


=  Fhy  =  7tr*hy 


=  *  8'  x 


and  is  applied  through  the  centre  =j*/_ 

of  gravity  of  the   circle.     Eoi-  'ffiw////////////w^^^ 

dently  R  is  the  same  for  any 


depth  below  the  tail-water  surface,  so  long  as  h  —  2  ft. 
the  student  find  a  graphic  proof  of  this  statement.] 


[Let 


444.  Liquid  Pressure  on  Curved  Surfaces.  —  If  the  rigid  surface 
is  curved,  the  pressures  on  the  individual  dJFy&,  or  elements  of 
area,  do  not  form  a  system  of  parallel  forces,  and  the  single  re- 
sultant (if  one  is  obtainable)  is  not  equal  to  their  sum.  In 


f)70 


MECHANICS   OF   ENGINEERING. 


FIG.  495. 


general,  the  system  is  not  equivalent  to  a  single  force,  but  can 
always  be  reduced  to  two  forces  (§  38)  the  point  of  application 
of  one  of  which  is  arbitrary  (the  arbitrary  origin  of  §  38)  and 
its  amount  =  V(2X)*  +  (2  Y)*  +  (2Z)\ 

A  single  Example  will  be  given  ;  that  of  a  thin  rigid  shell 
having  the  shape  of  the  curved  surface  of  a  right  cone,  Fig. 
495,  its  altitude  being  h  and  radius  of  base  =  r.  It  has  no 
bottom,  is  placed  on  a  smooth  horizontal  table,  vertex  up,  and 
is  filled  with  water  through  a  small  hole  in  the  apex  0,  which  is 

left  open  (to  admit  atmospheric 
pressure).  What  load,  besides  its 
own  wejght  O-'  ',  must  be  placed 
upon  it  to  prevent  the  water  from 
lifting  it  and  escaping  under  the 
edge  A  ?  The  pressure  on  each 
dF  of  the  inner  curved  surface  is 
is  normal  to  the  surface. 

. 

Its  vertical  cornpon.  is  zydffSLn  ot, 

and  horizontal  compon.  =•  zydF  cos  a.  The  dF's  have  all 
the  same  #,  but  different  z's  (or  heads  of  water).  The  lifting 
tendency  of  the  water  on  the  thin  shell  is  due  to  the  vertical 
components  forming  a  system  of  ||  forces,  while  the  horizon- 
tal components,  radiating  symmetrically  from  the  axis  of  the 
cone,  neutralize  each  other.  Hence  the  resultant  lifting  force 
is 

V  =  J£(vert.  comps.)  =  y  sin  afzdF-=.  y  sin  a  Fz  ;     (1) 

where  F  =  total  area  of  curved  surface,  and  z  =  the  "head  of 
water"  of  its  centre  of  gravity.  Eq.  (1)  may  also  be  written 
thus: 

V=yF~z;     .    .     .    '.     .     .     .     (2) 

in  which  Fb  =  7^  sin  a  =  area  of  the  circular  base  =  area  of 
the  projection  of  the  curved  surface  upon  a  plane  ~|  to  the 
vertical,  i.e.,  upon  a  horizontal  plane.  Hence  we  may  write 


(8) 


since  z  =  %h,  being  the  z  of  the  centre  of  gravity  of  the  curved 


CONICAL    SHELL.  571 

surface  and  not  that  of  the  base,  y  —  heaviness  of  water.  If 
G-'  —  weight  of  the  shell  and  is  <  "F,  an  additional  load  of 
V  —  G'  will  be  needed  to  prevent  the  lifting.  If  the  shell  has 
a  bottom  of  weight  =  G"  ^  forming  a  base  for  the  cone  and 
rigidly  attached  to  it,  we  find  that  the  vertical  forces  acting  on 
the  whole  rigid  body,  base  and  all,  are:  V upward;  G'  and 
G"  downward;  and  the  liquid  pressure  on  the  base,  viz., 
V  =  nr*hy  (§  428&)  also  downward.  Hence  the  resultant 
vertical  force  to  be  counteracted  by  the  table  is  downward,  and 

=  G'  +  G"  +  V  -  F,  which  =  G'  +  G"  +  Ixr'hr  5    W 

i.e.,  the  total  weight  of  the  rigid  vessel  and  the  water  in  it,  as 
we  know,  of  course,  in  advance. 


CHAPTEK  III. 

EARTH  PRESSURE  AND  RETAINING  WALLS. 


.  —  This  chapter  was  outlined  and  written  mainly  by 
Prof.  C.  L.  Crandall,  and  is  here  incorporated  with  his  permis- 
sion. The  theory  of  earth  pressure  is  arranged  from  Bau- 
meister.] 

445.  Angle  of  Repose.  —  Granular  materials,  like  dry  sand,. 
loose  earth,  soil,  gravel,  pease,  shot,  etc.,  on  account  of  the 
friction  between  the  component  grains,  occupy  an  intermediate 
position  between  liquids  and  large  rigid  bodies.  When  heaped 
up,  the  side  of  the  mass  cannot  be  made  to  stand  at  an  inclina- 
tion with  the  horizontal  greater  than  a  definite  angle  called  the 
angle  of  natural  slope,  or  angle  of  repose,  different  for  each 
material  ;  so  that  if  the  side  of  the  mass  is  to  be  retained  per- 
manently at  some  greater  angle,  a  Retaining  Wall  (or  "  Revet- 
ment Wall"  in  military  parlance)  becomes  necessary  to  sup- 
port it.  If  the  material  is  somewhat  moist  it  may  be  made  to 
stand  alone  at  an  inclination  greater  than  that  of  the  natural 
slope,  on  account  of  the  cohesion  thus  produced,  but  only  as 
long  as  the  degree  of  moisture  remains  ;  while  if  much  water 
is  present,  it  assumes  the  consistency  of  mud  and  may  require 
a  much  thicker  wall,  if  it  is  to  be  supported  laterally,  than  if 
dry. 

In  dealing  with  earth  to  be  supported  by  a  retaining  wall, 
we  consider  the  former  to  have  lost  any  original  cohesion 
which  may  have  existed  among  its  particles,  or  that  it  will 
eventually  lose  it  through  the  action  of  the  weather  ;  and  hence 
treat  it  as  a  granular  material. 

A  few  approximate  values  of  the  angle  of  natural  slope  are 

572 


RETAINING    WALLS. 


573 


given  below,  being  taken  from  Fanning,  p.  345 ;  see  reference 
on  p.  538  of  this  work. 


MATERIAL. 

Angle 
of  Repose. 

Coefficient 
of  Friction. 

Ratio 
of  Slope. 

Dry  suiid    fine             

28° 

.532 

Horiz.  to  vert. 
1.88  to    1 

"       "      coarse  

30° 

.577 

1.73          1 

45° 

1.000 

1.00          1 

Wet  clay 

15° 

.268 

3.73          1 

Clayey  gravel  

45° 

1.000 

1.00         1 

42° 

.900 

1.11          1 

Gravel                        

38° 

.781 

1.28          1 

Firm  loam  

36° 

.727 

1.38          1 

35° 

.700 

1.43         1 

Peat                   

20° 

.364 

2.75         1 

The  angle  of  repose,  or  natural  slope,  is  also,  evidently,  the 
angle  of  friction  between  two  masses  of  the  same  granular 
material. 

446.  Earth  Pressure,  and  Wedge  of  Maximum  Thrust. — Fig. 
496.  Let  AB  be  a  retaining  wall,  having  a  plane  face  AB  in 
contact  with  a  mass  of  earth  ABD,  both  wall  and  earth  being 
of  indefinite  extent  ~|  to  the  paper. 

Let  AD  be  the  natural  slope  of  the  earth,  making  an  angle  ft 
with  the  vertical  (ft  is  the  complement  of  the  angle  of  repose ; 
see  preceding  table).  Since  AB,  making  an  angle  a  with 
the  vertical,  is  more  nearly  vertical  than  AD,  the  retaining 
wall  is  necessary,  to  keep  the  mass  ABD  in  the  position 
shown.  The  profile  BCD  may  be  of  any  form  in  this  general 
discussion.  Suppose  the  wall  to  be  on  the  point  of  giving 
way  ;  then  the  following  motions  are  impending  : 

1st.  Sliding  is  impending  between  some  portion  AB C' A  of 
the  mass  of  earth  and  the  remainder  C' AD,  the  surface  of 
rupture  AC'  (CJ  not  shown  in  figure  because  not  found  yet, 
but  lying  somewhere  on  the  profile  BCID)  being  assumed 
plane,  and  making  some  angle  0'  (to  be  determined)  with  the 
vertical.  At  this  instant  the  resultant  pressure  N'  of  AC'D 
on  the  plane  AC'  of  the  mass  ABC'  (a  wedge)  must  make 
an  angle  =  ft  (=  comp.  of  angle  of  friction)  with  AC'  on 
the  upper  side. 


574 


MECHANICS    OF    ENGINEERING. 


2d.  A  downward  sliding  of  the  mass  ABC'  along  the  back 
face  AB  of  the  wall.     That  is,  the  resultant  pressure  P'  of 
the  wall  against  the  mass  BA  C'  at  this  instant  makes  an  angle 
y 

I  \       EARTH  SURFACE  .^H 

I A 


S  (=  complement  of  angle  of  friction  between  the  earth  and 
wall)  with  the  plane  AB  and  on  the  upper  side.  The  weight 
of  the  wedge  of  earth  BAC'  will  be  called  &',  and  we  desire 
to  find  the  pressure  P'  against  the  wall. 

Let  BAC  be  a  wedge^(of  the  earth-mass),  in  which  AC  makes 
any  angle  0  with  Af,  and  suppose  it.  to  be  on  the  point  of 
moving  down  and  forcing  out  the  wall;  thus  encountering 
friction  both  on  the  plane  A  C  and  the  plane  AB.  Then  the 
forces  acting  on  it  are  three,  acting  in  known  directions  ;  viz. : 
G,  its  own  weight,  vertical ;  N^  the  resultant  pressure  of  the 
earth  below  it,  making  an  angle  fi  with  AC  on  upper  side  ; 
and  P,  the  resultant  pressure  of  the  wall,  at  angle  6  with  AB 
(see  Fig.  496  for  positions  of  JV  and  P).  If  now  we  express 
the  force  P  in  terras  of  0  and  other  quantities,  and  find  that 
value  0',  of  0,  for  which  P  is  a  maximum,  we  thereby  deter- 
mine the  "wedge  of  maximum  thrust"  ABC' A  ;  while  this 
maximum  thrust,  P',  is  the  force  which  the  wall  must  be  de- 
signed to  withstand.  [If  the  wall  is  overturned,  the  earth 
will  sink  with  it  until  this  part  of  its  surface  gradually  as- 
sumes the  natural  slope.] 

Let  Gr  =  weight  of  prism  of  base  ABC,  and  altitude  =  unity 
~1  to  paper;  then  G  =  y  X  area  ABC,  where  y  =  "  heavi- 
ness" =  wgt.  per  cub.  unit,  of  earth.  Now  P,  G,  and  N 
balance;  therefore,  in  triangle  abc,  if  ab  and  ac  are  drawn  j| 


RETAINING   WALLS. 


and  —  G  and  .^respectively,  be  is  =  and  ||  to  P;  and  from 
Trigonometry  we  have 

p_      ^in[/?-0]      m 

' 


in  which  tf  stands  for  a  +  0,  for  brevity,  being  the  angle 
which  P  makes  with  the  vertical.  N  makes  an  angle  =  ft  —  0 
with  the  vertical. 

The  value,  0',  of  0,  which  makes  P  a  maximum  is  found 

by  placing  ——  =  0.  From  eq.  (1),  remembering  that  G  is  a 
function  of  0,  and  that  f3  and  $  are  constants,  we  have 

sin(/3  +  S-<f>)j_-sin(/3-(!>)  -  G  cos  (|3  -  </>)~|  -f£sin(|3  -  <f>)  cos  (j3  +  8  -  <£) 
dtfT  ~  sln2  [j8  -f  6  -  <f>] 

For  P  to  be  a  maximum  we  must  put 

numerator  of  above  —  0 (#} 


To  find  a  geometrical  equivalent  of  — ,  denote  A  C  by  Z, 

and  draw  AE,  making  an  angle  =  d<p  with  AC.  Now  the 
area  ACI '=  37  X  %(fE=(L  +  dL)^Ld(f>  =  %Z?d(j)  .  .  . 
(neglecting  infinitesimal  of  2d  order).  Now 

dG  =  y  X  area  J. CI X  unity ;  /.  -=--  =  i/Z2;  /.  (a)  becomes 


sin(/3  +  d—  0)^Z3sin(/?—  0)—  sin  (/?  +  (J  —  0)^cos(^—  0) 

+  #  sin  (/?  —  0)  cos  (/?  +  d  —  0)  =  0  ; 

i.e.,  G  = 

sin  (ft  —  0)  sin  (ft  +  £  —  0)  _ 


sin  (/?  -f  d  —  0)  cos  (/?  —  0)  —  cos  (yff  +  d  —  0)  sin  (/?  —  0) 

-    ./ 


576  MECHANICS   OF   ENGINEERING. 

when  P  is  a  maximum  ;  and  hence,  calling  G-'  and  0'  and  Z' 
the  values  of  6r,  0,  and  Z,  for  max.  P,  we  have 


in  (ft  -  ,   •    •    (2) 


and  therefore  from  (1)  P  max.  itself  is 


447.  Geometric  Interpretation  and  Construction.  —  If  in  Fig. 
496  we  draw  CF^  making  angle  d  with  AD,  C  being  any 
point  on  the  ground  surface  JBD,  we  have 


sn 
Drop  a  perpendicular  FH  from  F  to  A  C,  and  we  shall  have 

=  CF.  sin  (ft  +  S  -  <f\  =  L  .  ™(fi 


From  this  it  follows  that  the  weight  of  prism  of  base  A  CF 
and  unit  height 


When  AC  (as  0  varies)  assumes  the  position  and  value  AC  ', 
bounding  the  prism  of  maximum  thrust,  Fig.  497,  Z  becomes 
=  Zr,  and  0  =  0r;  and  eq.  (4)  gives  the  weight  of  the  prism 
AC  Fr.  This  weight  is  seen  to  be  equal  to  that  of  the  prism 
(or  wedge)  of  maximum  thrust  ABC',  by  comparing  eq.  (4) 
with  eq.  (2);  that  is,  AC'  Usects  the  area  ABC'F',  and 
hence  may  be  determined  by  fixing  such  a  point  <7',  on  the 
upper  profile  BD,  as  to  make  the  triangular  area  AC'F' 
equal  to  the  sectional  area  of  the  wedge  BC'A  ;  C'F'  being 
drawn  at  an  angle  =  d  with  AD. 

This  holds  for  any  form  of  ground  surface  BD,  or  any 


KETAININO    WALLS. 


577 


values  of  the  constants  /?,  a,  or  6.     Cf  is  best  found  graphic- 
ally by  trial,  in  dealing  with 
an  irregular  profile  BD. 

Having  found  AC  ',  = 
Z',  P'  can  be  found  from 
(3),  or  graphically  as  fol- 
lows :  (Fig.  497)  With  F' 
as  a  centre  and  radius  = 
C'F')  describe  an  arc  cut- 
ting AD  in  J',  and  join 
C'J'.  The  weight  of  prism  FIG.  497. 

with  base  CltL'F'  and  unit  height  will  =  P'.    For  that  prism 
has  a  weight 


;  but 


sin 


but 
and 


sind 


'.#'  =  Z'  sin  08  - 


.-.      weight  of  prism  C'J'F  =  ±yL' 

[See  eq.  (3).] 


sin  o 


448.  Point  of  Application  of  the  Resultant  Earth  Thrust.— 
This  thrust  (called  Pf  throughout  this  chapter  except  in  the 
present  paragraph)  is  now  known  in  magnitude  and  direction, 
but  not  in  position  ;  i.e.,  we  must  still  determine  its  line  of 
action,  as  follows  : 

Divide  AB  into  a  number  of  equal  parts,  ab,  ~bc,  cd,  etc.; 
see  Fig.  498.  Treat  ab  as  a  small  retaining  wall,  and  find  the 
magnitude  P'  of  the  thrust  against  it  by  §  447  ;  treat  ac  simi- 
larly, thus  finding  the  thrust,  P"  ^  against  it  ;  then  ad,  ae,  etc., 
the  thrusts  against  them  being  found  to  be  P"  ',  P^,  etc.  ;  and 
so  on.  Now  the  pressure 

P'  on  ab  is  applied  nearly  at  middle  of  ob, 
P"  -  P'          «  «  "        «     Ic, 


578  MECHANICS   OF  ENGINEERING. 

and  so  on.  Erect  perpendiculars  at  tlie  middle  points  of  cib, 
Ic,  cd,  etc.,  equal  respectively  to  P\ 
P"  -  Pf,  P'"  -  P",  etc.,  and  join  the 
ends  of  the  perpendiculars.  The  per- 
pendicular through  the  centre  of  gravity 
of  the  area  so  formed  (Fig.  498)  will 
give,  on  AB,  the  required  point  of  ap- 
plication of  the  thrust  or  earth  pressure 
on  AB,  and  this,  with  the  direction  and 
FIG.  498.  magnitude  already  found  in  §  447,  will 

completely  determine  the  thrust  against  the  wall  AB. 

449.  Special  Law  of  Loading. — If  the  material  to  be  retained 
consists  of  loose  stone,  masses  of  masonry,  buildings,  or  even 
moving  loads,  as  in  the  case  of  a  wharf  or  roadway,  each  can 
be  replaced  by  the  same  weight  of  earth  or  other  material 
which  will  render  the  bank  homogeneous,  situated  on  the  same 
verticals,  and  the  profile  thus  reduced  can  be  treated  by  §§  447 
and  448. 

Should  the  solid  mass  extend  below  the  plane  of  rupture, 
AC',  and  the  plane  of  natural  slope,  it  will  become  a  retaining 
wall  for  Ihe  material  beyond,  if  strong  enough  to  act  as  such 
(limiting  the  profile  ABCD  of  Fig.  496  to  the  front  of  the 
mass,  or  to  the  front  and  line  of  rupture  for  maximum  thrust 
above  it,  if  it  does  not  reach  the  surface);  if  not  strong  enough, 
or  if  it  does  not  reach  below  the  plane  of  natural  slope,  it& 
presence  is  better  ignored,  probably,  except  that  the  increased 
weight  must  be  considered. 

The  spandrel  wall  of  an  arch  may  present  two  of  these 
special  cases ;  i.e.,  the  profile  may  be  enlarged  to  include  a 
moving  load,  while  it  may  be  limited  at  the  back  by  the  other 
spandrel. 

If  the  earth  profile  starts  at  the  front  edge  of  the  top  of 
wall,  instead  of  from  the  back  as  at  B,  Fig.  496,  eq.  (3)  would 
only  apply  to  the  portion  behind  AB  prolonged,  leaving  the 
part  on  the  wall  (top)  to  be  treated  as  a  part  of  the  wall  to  aid 
in  resisting  the  thrust. 

If  the  wall  is  stepped  in  from  the  footings,  or  foundation 


RETAINING    WALLS.  579 

courses,  probably  the  weak  section  will  be  just  above  them ;  if 
stepped  at  intervals  up  the  back  of  the  wall,  the  surface  of  separa- 
tion between  the  wall  and  filling,  if  it  is  plane,  will  probably 
pass  through  the  first  step  and  incline  forward  as  much  as  pos- 
sible without  cutting  the  wall. 

450.  Straight  Earth-profile, — The  general  case  can  be  simpli- 
fied as  follows  (the  earth-profile  BD  being  straight,  at  angle 
=  £  with  vertical,  =  DET) :  Since  the  triangles'  ABC'  and 


C'AF'  are  equal,  from  §  447,  and  AC'  is  common,  therefore 
BS=F'H  (both  being  drawn  1  to  AC').  Draw  AE  and 
BM  ||  to  F'C'  (i.e.,  at  angle  d  with  AD\  cutting  DB,  pro- 
longed, in  E.  We  have 

DE  EA  ,  CHE  EA 

,   and 


G'E       EA  -  G'F' BE       EA- 


But  C'F'  =  BM  (since  BS  =  H'F1)  ; 


therefore  =    ~;  i.e.,  DE  .  BE  =  C'E\ 

C'E       BE 

which  justifies  the  following  construction  for  locating  the  de- 
sired point  C'  on  BD,  and  thus  finding  AC'  —  L'  and  the 
angle  <f>'\  Describe  a  circle  on  ED  as  a  diameter,  and  draw 


580  MECHANICS   OF   ENGINEERING. 

BX  ~l  to  BD,  thus  fixing  X  in  the  curve.  With  centre  E 
describe  a  circular  arc  through  X,  cutting  BD  in  C',  required. 

Having  AC'  (i.e.,  L'\  0'  is  known  ;  hence  from  eq.  (3)  we 
obtain  the  earth  thrust  or  pressure  P'\  or,  with  F1  as  centre 
and  radius  =  C'F',  describe  arc  C'J'\  then  the  triangle  C'F'J' 
is  the  base  of  a  prism  of  unity  height  whose  weight  =  P'  (as 
in  §  447). 

Centre  of  Pressure. — Applying  the  method  of  §  448,  Fig. 
498,  to  this  case,  we  find  that  the  successive  L'  's  are  propor- 
tional to  the  depths  ab,  ac,  ad,  etc.,  and  that  the  successive  P'B 
are  proportional  [see  (3)]  to  the  squares  of  the  depths ;  hence 
the  area  in  Fig.  498  must  be  triangular  in  this  case,  and  the 
point  of  application  of  the  resultant  pressure  on  AB  is  one 
third  of  AB  from  A  :  just  as  with  liquid  pressure. 

451.  Resistance  of  Retaining  Walls.— (Fig.  500.)  Knowing 
the  height  of  the  wall  we  can  find  its  weight,  =  Gl ,  for  an  as- 
sumed thickness,  and  unity  width  ~|  to  paper.  The  resultant 
of  Cr1 ,  acting  through  the  centre  of  gravity  of  wall,  and  P\  the 
thrust  of  the  embankment,  in  its  proper 
line  of  action,  should  cut  the  base  A  V 
within  the  middle  third  and  make  an 
angle  with  the  normal  (to  the  base)  less 
than  the  angle  of  friction. 

For  the  straight  wall  and  straight 
earth-profile  of  Fig.  499  and  §  450,  the 
FIG.  500.  length  Z',  =  A  C ',  can  be  expressed  in 

terms  of  the  (vertical)  height,  A,  of  wall,  thus : 


'  =  AC'  =  B{n  (C~  g)  sin  (C  ~  a 


-      —  . 
sin  (C  —  0  )        cos  a     sin  (£  —  0  ) 

.-.  eq.  (3)  becomes 

P'  =  lr  Jt-     Bin°(/?-0Qsin*(C-«)  K 

^r  cos2  a   '        sin  d  sin2(C  -  00  cosa  a 

[A  representing  the  large  fraction  for  brevity.] 


RETAINING    WALLS. 


581 


This  equation  will  require,  for  a  wall  of  rectangular  section, 
that  the  thickness,  d,  increase  as  A,  in  order  that  its  weight  may 
increase  as  A3  (i.e.,  as  Pf)  and  that  its  resisting  moment  may- 
increase  with  the  overturning  moment. 

By  this  equality  of  moments  is  meant  that  P'a  =  Gfi ; 
where  a  and  b  are  the  respective  lever-arms  of  the  two  forces 
about  the  front  edge  of  the  middle  third.  (AB  is  the  back  of 
the  wall.)  In  other  words,  their  resultant  will  pass  through 
this  point. 

The  following  table  is  computed  on  the  basis  just  mentioned, 
viz.,  that  the  resultant  of  P'  and  G-  shall  pass  through  the 
front  edge  of  the  midde  third. 

The  wall  is  vertical,  i.e.,  a  is  =  0,  and  is  of  rectangular  sec- 
tion ;  and  we  further  suppose  that  the  heaviness  of  the  earth  is 
two  thirds  that  of  the  masonry  of  the  wall ;  d  is  tne  proper 
safe  thickness  to  be  given  to  the  wall  on  the  basis  spoken  of,  h 
being  its  altitude.  Whether  the  wall  is  safe  against  sliding  on 
its  base,  and  whether  a  safe  compression  per  unit  area  is  ex- 
ceeded on  the  front  edge  of  the  base,  are  matters  for  separate 
consideration.  See  Figs.  499  and  500,  and  the  foregoing  text, 
for  the  meaning  of  all  symbols  employed.  The  above  assump- 
tion as  to  the  relative  densities  of  wall  and  earth  is  realized  if 
the  wall  is  of  first-quality  masonry  weighing  150  Ibs.  per  cubic 
foot,  supporting  earth  of  100  Ibs.  per  cubic  foot.  Note  that 
d  —  a  +  8 ;  i.e.,  d  —  8  f or  this  table. 

a  =  0;  i.e.,  wall  is  vertical;  also  density  of  wall  =  f  that  of  the  earth. 


I. 

II. 

III. 

<J  =  90° 
0  =  90° 

i;r 

is« 

tan/3 

ft 

<*>' 

A 

d 

f 

A 

d 

*' 

A 

d 

1.0 

45° 

22^° 

.17 

Mh 

26° 

.18 

.22* 

45° 

.71 

.33* 

1.5 

56i° 

28° 

.29 

Mli 

33° 

.26 

.30* 

56° 

.83 

,43* 

2.0 

63|° 

31|° 

.38 

.51* 

38° 

.33 

.36* 

63° 

.89 

.51* 

4.0 

76° 

38° 

.61 

.64* 

45° 

.54 

.50* 

76° 

.97 

.65* 

Infinity 

90° 

45° 

1.00 

.82* 

90° 

1.00 

.82* 

90° 

1.00 

.82* 

In  Case  I  of    table,  since   a  =  0,  0  =  90°  and  C  =  90° ; 
d  =  90°,  and  hence  C'F'  of  Fig.  499  is  1  to  AD,  so  that 


582 


MECHANICS   OF   ENGINEEKING. 


(since  the    area    of    &ABC'  =  &ACfFf)    0'  must  =  fyff. 
These  values,  in  (5),  give 

P'  =  %yh*  tan2  %ft ;  i.e.,  A  =  tan2  \ft.  .    .     .    (6) 

In  Case  II,  since  C  =  90°,  a  =  0  and  8  =  ft,    .-.    3  =  ft ; 
and  (5)  reduces  to 

;in2  (ft  —  (f>'\    .         .  sin2  (ft  —  0r)          ,^. 


In  Case  III,  C  =  /?  and 


will  be  ||  to  J.Z>,  Z>  being  at 
infinity.    See  Fig.  501.   Through 


FIG.  SOL 


BI1  1  to  u4Z>,  and  BF" 
e  d  with  ^LI>.  C'  is 
to  be  located  on  BD,  so  as 
tQ  make  (area  of)  &ABC'  = 
(area  of)  A  AC'f"  (according 
to  §  447),  the  angle  C  'F'A  being 
=  <$  =  a+  Q\  =  0,  in  this  case, 
and  hence  also  =  ft.  Conceive 
B  and  F  '  to  be  joined. 


Now 


'F'  =  &ABF"  + 


But  &ABC'   =  kBF'F"  (equal  bases  and  altitudes). 

Hence  A  ABC'  cannot  =  i^AC'F'  unless  Cf  is  moved  out 
to  infinity  ;  and  then  0'  becomes  —  ytf,  and  eq.  (5)  reduces  to 


P'=  ^ytf  sin  ft ;    i.e.,  A  =  sin  ft. 


(8) 


[Increasing  «  from  zero  will  decrease  the  thickness  d ;  i.e., 
inclining  the  wall  inwards  will  decrease  the  required  thickness, 
but  diminish  the  frictional  stability  at  the  base,  unless  the  lat- 
ter be  "1  to  AB.  The  back  of  the  wall  is  frequently  inclined 
outwards,  making  the  section  a  trapezoid,  to  increase  the  fric- 
tional stability  at  the  base  when  necessary,  as  with  timber 
walls  supporting  water.] 


KETAINLNG    WALLS.  583 

452.  Practical  Considerations.  —  An  examination  of  the 
values  of  A  and  d  in  the  table  of  §  451  will  show  that  in  sup- 
porting quicksand  and  many  kinds  of  clay  which  are  almost 
fluid  under  the  influence  of  water,  it  is  important  to  know 
what  kind  of  drainage  can  be  secured,  for  on  that  will  depend 
the  thickness  of  the  wall.  With  well  compacted  material  free 
from  water-bearing  strata,  an  assumed  natural  slope  of  If  to  1 
(i.e.,  If  hor.  to  1  vert.)  will  be  safe ;  the  actual  pressure  below 
the  effect  of  frost  and  surface  water  will  be  that  due  to  a  much 
steeper  slope  on  account  of  cohesion  (neglected  in  this  theory). 

The  thrust  from  freshly  placed  material  can  be  reduced  by 
depositing  it  in  layers  sloping  back  from  the  wall.  If  it  is  not 
so  placed,  however,  the  natural  slope  will  seldom  be  flatter 
than  If  to  1  unless  reduced  by  water.  In  supporting  material 
which  contains  water-bearing  strata  sloping  toward  the  wall 
and  overlain  by  strata  which  are  liable  to  become  semi-fluid 
and  slippery,  the  thrust  may  exceed  that  due  to  semi-fluid  ma- 
terial on  account  of  the  surcharge.  If  these  strata  are  under 
the  wall  and  cannot  be  reached  by  the  foundation,  or  if  resist- 
ance to  sliding  cannot  be  obtained  from  the  material  in  front 
by  sheet-piling,  no  amount  of  masonry  can  give  security. 

Water  at  the  back  of  the  wall  will,  by  freezing,  cause  the 
material  to  exert  an  indefinitely  great  pressure,  besides  disinte- 
grating the  wall  itself.  If  there  is  danger  of  its  accumulation, 
drainage  should  be  provided  by  a  layer  of  loose  stone  at  the 
back  leading  to  "weep-holes"  through  the  wall. 

A  friction-angle  at  the  back  of  the  wall  equal  to  that  of  the 
filling  should  always  be  realized  by  making  the  back  rough  by 
steps,  or  projecting  stones  or  bricks.  Its  effect  on  the  required 
thickness  is  too  great  to  be  economically  ignored. 

The  resistance  to  slipping  at  the  base  can  be  increased,  when 
necessary,  by  inclining  the  foundation  inwards;  by  stepping 
or  sloping  the  back  of  the  wall  so  as  to  add  to  its  effective 
weight  or  incline  the  thrust  more  nearly  to  the  vertical;  by 
sheet-piling  in  front  of  the  foundation,  thus  gaining  the  resist- 
ance offered  by  the  piles  to  lateral  motion  ;  by  deeper  founda- 
tions, gaining  the  resistance  of  the  earth  in  front  of  the  wall. 


584  MECHANICS    OF   ENGINEERING. 

The  coefficient  of  friction  on  the  base  ranges,  according  to 
Trautwine,  from  0.20  to  0.30  on  wet  clay ; 
"       .50  to    .66   "  dry  earth; 
"       .66  to    .75    "  sand  or  gravel ; 
"       .60  on  a  dry  wooden  platform  ;  to  .75  on  a 
wet  one. 

If  the  wall  is  partially  submerged,  the  buoyant  effort  should 
be  subtracted  from  G1 ,  the  weight  of  wall. 

453.  Eesults  of  Experience. — (Trautwine.)  In  railroad  prac- 
tice, a  vertical  wall  of  rectangular  section,  sustaining  sand, 
gravel,  or  earth,  level  with  the  top  [p.  682  of  Civ.  Eng.  Pocket 
Book]  and  loosely  deposited,  as  when  dumped  from  carts,  carsr 
etc.,  should  have  a  thickness  d,  as  follows : 

If  of  cut  stone,  or  of  first-class  large  ranged  rubble,  in  mortar.  .  .  .  d  =  .  35& 

"    good  common  scabbled  mortar-rubble,  or  brick d  —  AOh 

"    well  scabbled  dry  rubble d  =  .5Qh 

Where  h  includes  the  total  height,  or  about  3  ft.  of  foundations. 

(a)  For  the  best  masonry  of  its  class  h  may  be  taken  from 
the  top  of  the  foundation  in  front. 

(5)  A  mixture  of  sand  or  earth,  with  a  large  proportion  of 
round  boulders  or  cobbles,  will  weigh  more  than  the  backing 
assumed  above  ;  requiring  d  to  be  increased  from  one  eighth  to 
one  sixth  part. 

(<?)  The  wall  will  be  stronger  by  inclining  the  back  inwards, 
especially  if  of  dry  masonry,  or  if  the  backing  is  put  in  place 
before  the  mortar  has  set. 

(d)  The  back  of  the  wall  should  be  left  rough  to  increase 
friction. 

(e)  Where  deep  freezing  occurs,  the  back  should  slope  out- 
ward for  3  or  4  feet  below  the  top  and  be  left  smooth. 

(«/)  When  a  wall  is  too  thin,  it  will  generally  fail  by  bulging 
outward  at  about  one  third  the  height.  The  failure  is  usually 
gradual  and  may  take  years. 

(g)  Counterforts,  or  buttresses  at  the  back  of  the  wall,  usually 
of  rectangular  section,  may  be  regarded  as  a  waste  of  ma- 
sonry, although  considerably  used  in  Europe ;  the  bond  will 


RETAINING    WALLS.  585 

seldom  hold  them  to  the  wall.  Buttresses  in  front  add  to  the 
strength,  but  are  not  common,  on  account  of  expense. 

(A)  Land-ties  of  iron  or  wood,  tying  the  wall  to  anchors  im- 
bedded below  the  line  of  natural  slope,  are  sometimes  used  to 
increase  stability. 

(i)  Walls  with  curved  cross-sections  are  not  recommended. 

454.  Conclusions  of  Mr.  B.  Baker. — ("  Actual  Lateral  Pressure 
of  Earthwork.")     Experience  has  shown  that  d  =  0.25 A,  with 
batter  of  1  to  2  inches  per  foot  on  face,  is  sufficient  when 
backing  and  foundation  are  both  favorable ;  also  that  under  no 
ordinary  conditions  of  surcharge  or  heavy  backing,  with  solid 
foundation,  is  it  necessary  for  d  to  be  greater  than  0.50A. 

Mr.  Baker's  own  rule  is  to  make  d  =  0.33A  at  the  top  of 
the  footings,  with  a  face  batter  of  If  inches  per  foot,  in  ground 
of  average  character ;  and,  if  any  material  is  taken  out  to  form 
a  face-panel,  three  fourths  of  it  is  put  back  in  the  form  of  a 
pilaster.  The  object  of  the  batter,  and  of  the  panel  if  used,  is 
to  distribute  the  pressure  better  on  the  foundation.  All  the 
walls  of  the  "  District  Railway"  (London)  were  designed  on 
this  basis,  and  there  has  not  been  a  single  instance  of  settle- 
ment, of  overturning,  or  of  sliding  forward. 

455.  Experiments  with  Models. — Accounts   of   experiments 
with  apparatus  on  a  small  scale,  with  sand,  etc.,  may  be  found 
in  vol.  LXXI  of  Proceedings  of  Institution  of  Civil  Engineers, 
London,  England  (p.  350) ;  also  in  vol.  n  of  the  "  Annales  des 
Ponts  et  Chaussees"  for  1885  (p.  788). 


CHAPTEE  IY. 

HYDROSTATICS  (Continued)— IMMERSION  AND  FLOTATION. 

456.  Rigid  Body  Immersed  in  a  Liquid.  Buoyant  Effort. — If 
any  portion  of  a  body  of  homogeneous  liquid  at  rest  be  con- 
ceived to  become  rigid  without  alteration  of  shape  or  bulk,  it 
would  evidently  still  remain  at  rest ;  i.e.,  its  weight,  applied  at 
its  centre  of  gravity,  would  be  balanced  by  the  pressures,  on  its 
bounding  surfaces,  of  the  contiguous  portions  of  the  liquid ; 
hence, 

If  a  rigid  body  or  solid  is  immersed  in  a  liquid,  ~boih  ~being 
at  rest,  the  resultant  action  upon  it  of  the  surrounding  liquid 
(or  fluid)  is  a  vertical  upward  force  called  the  "buoyant 
effort"  equal  in  amount  to  the  weight  of  liquid  displaced, 
and  acting  through  the  centre  of  gravity  of  the  volume  (con- 
sidered as  homogeneous)  of  displacement  (now  occupied  by  the 
solid).  This  point  is  called  the  centre  of  buoyancy,  and  is 
sometimes  spoken  of  as  the  centre  of  gravity  of  the  displaced 
water.  If  V  =  the  volume  of  displacement,  and  y  =  heavi- 
ness of  the  liquid,  then  the 

buoyant  effort  ==  V'y (1) 

(By  "  volume  of  displacement"  is  meant,  of  course,  the  volume 
of  liquid  actually  displaced  when  the  body  is  immersed.) 

If  the  weight  G1  of  the  solid  is  not  equal  to  the  buoyant 
effort,  or  if  its  centre  of  gravity  does  not  lie  in  the  same  verti- 
cal as  the  centre  of  buoyancy,  the  two  forces  form  an  unbal- 
anced system  and  motion  begins.  But  as  a  consequence  of 
this  very  motion  the  action  of  the  liquid  is  modified  in  a  man- 
ner dependent  on  the  shape  and  kind  of  motion  of  the  body. 

586 


IMMERSION. 


587 


Problems  in  this  chapter  are  restricted  to  cases  of  rest,  i.e., 
balanced  forces. 

Suppose  G'  —  V'y ;   then, 

If  the  centre  of  gravity  lies  in  the  same  vertical  line  as  the 
centre  of  buoyancy  and  underneath  the  latter,  the  equilibrium 
is  stable ;  i.e.,  after  a  slight  angular  disturbance  the  body  re- 
turns to  its  original  position  (after  several  oscillations) ;  while 
if  above  the  latter,  the  equilibrium  is  unstable.  If  they  coin- 
cide, as  when  the  solid  is  homogeneous  (but  not  hollow),  and 
of  the  same  heaviness  (§  7)  as  the  liquid,  the  equilibrium  is 
indifferent,  i.e.,  possible  in  any  position  of  the  body. 

The  following  is  interesting  in  this  connection : 

In  an  account  of  the  new  British  submarine  boat  "  JSTautilus," 
a  writer  in  Chambers*  8  Journal  remarked  [1887]  :  "At  each 
side  of  the  vessel  are  four  port-holes,  into  which  fit  cylinders 
two  feet  in  diameter.  When  these  cylinders  are  projected 
outwards,  as  they  can  be  by  suitable  gearing,  the  displacement 
of  the  boat  is  so  much  increased  that  the  vessel  rises  to  the 
surface;  but  when  the  cylinders  are  withdrawn  into  their 
sockets,  it  will  sink." 

As  another  case  in  point,  large  water-tight  canvas  "air-bags" 
have  recently  been  used  for  raising  sunken  ships.  They  are 
sunk  in  a  collapsed  state,  attached  by  divers  to  the  submerged 
vessel,  and  then  inflated  with  air  from  pumps  above,  which  of 
course  largely  augments  their  displacement  while  adding  no 
appreciable  weight. 

457.  Examples  of  Immersion. — Fig.  502.  At  (a)  is  an  ex- 
ample of  stable  equi- 
librium, the  centre  of 
buoyancy  B  being  above 
the  centre  of  gravity  (7, 
and  the  buoyant  effort 

V'y  —  G'  =  the  weight 
of  the  solid  ;  at  (a'\  con- 
versely, we  have  un- 
stable equilibrium,  with 

V'y  still  =  G'.     At  (1)  the  buoyant  effort  V'y  is  >  G',  and 


FIG.  502.] 


588  MECHANICS    OF   ENGINEERING. 

to  preserve  equilibrium  the  body  is  attached  by  a  cord  to  the 
bottom  of  the  vessel.  The  tension  in  this  cord  is 

•%>=  Vfy-Gf (1) 

At  (c)  V'y  is  <  G\  and  the  cord  must  be  attached  to  a 
support  above,  and  its  tension  is 

SC=G'-   V'y.      .     ......    v    (2) 

If  in  eq.  (2)  [(<?)  in  figure]  we  call  Sc  the  apparent  weight  of 
the  immersed  body,  and  measure  it  by  a  spring-  or  beam-bal- 
ance, we  may  say  that 

The  apparent  weight  of  a  solid  totally  immersed  in  a  liquid 
equals  its  real  weight  diminished  by  that  of  the  amount  of 
liquid  displaced  /  in  other  words,  the  loss  of  weight  —  the 
weight  of  displaced  liquid. 

EXAMPLE  1. — How  great  a  mass  (not  hollow)  of  cast-iron  can 
be  supported  in  water  by  a  wrought-iron  cylinder  weighing 
140  Ibs.,  if  the  latter  contains  a  vacuous  space  and  displaces 

3  cub.  feet  of  water,  both  bodies  being  completely  immersed  ? 
[Ft.,  lb.,  sec.] 

The  buoyant  effort  on  the  cylinder  is 

V'y  =  3  X  62.5  =  187.5  Ibs., 

leaving  a  residue  of  47.5  Ibs.  upward  force  to  buoy  the  cast- 
iron,  whose  volume  V"  is  unknown,  while  its  heaviness  (§  7) 
is  y"  =  450  Ibs.  per  cub.  foot.  The  direct  buoyant  effort  of 
the  water  on  the  cast-iron  is  V"y  —  [F"X62.5]  Ibs., 
and  the  problem  requires  that  this  force  +  47.5  Ibs.  shall 
=  V'y"  =  the  weight  G"  of  the  cast-iron ; 

.-.      V"  X  62.5  +  47.5  =  V"  X  450  ; 

/.    V"  —  0.12  cub.  ft.,  while  0.12  X  450=  54  Ibs.  of  cast-iron. 

Ans. 

EXAMPLE  2. — Required  the  volume  V,  and  heaviness  y'9 
of  a  homogeneous  solid  which  weighs  6  Ibs.  out  of  water  and 

4  Ibs.  when  immersed  (apparent  weight)  (ft.,  lb.,  sec.). 


IMMERSION.  589 

From  eq.  (2),  4  =  6  -  V  X  62.5  ;     .-.  V  —  0.032  cub.,  feet  ; 
....  y>  =  (}'  +  V  =  6  -=-  0.032  =  187.5  Ibs.  per  cub.  ft., 

and  the  ratio  of  y'  to  y  is  187.5  :  62.5  =  3.0  (abstract  num- 
ber) ;  i.e.,  the  substance  of  this  solid  is  three  times  as  dense, 
or  three  times  as  heavy,  as  water.  [The  buoyant  effort  of  the 
air  has  been  neglected  in  giving  the  true  weight  as  6  Ibs.] 

458.  Specific  Gravity.  —  By  specific  gravity  is  meant  the  ratio 
of  the  heaviness  of  a  given  homogeneous  substance  to  that  of 
a  standard  homogeneous  substance  ;  in  other  words,  the  ratio 
of  the  weight  of  a  certain  volume  of  the  substance  to  the 
weight  of  an  equal  volume  of  the  standard  substance.  Dis- 
tilled water  at  the  temperature  of  maximum  density  (4°  Centi- 
grade) under  a  pressure  of  147  Ibs.  per  sq.  inch  is  sometimes 
taken  as  the  standard  substance,  more  frequently,  however,  at 
62°  Fahrenheit  (16°.6  Centigrade).  Water,  then,  being  the 
standard  substance,  the  numerical  example  last  given  illustrates 
a  common  method  of  determining  experimentally  the  specific 
gravity  of  a  homogeneous  solid  substance,  the  value  there  ob- 
tained being  3.  The  symbol  cr  will  be  used  to  denote  specific 
gravity,  which  is  evidently  an  abstract  number.  The  standard 
substance  should  always  be  mentioned,  and  its  heaviness  y  ; 
then  the  heaviness  of  a  substance  whose  specific  gravity  is  cr  is 


(i) 


and  the  weight  G'  of  any  volume  V  of  the  substance  may  be 
written 

G1  =  Vy  =  V'<ry.     ......     (2) 

Evidently  a  knowledge  of  the  value  of  y'  dispenses  with  the 
use  of  cr,  though  when  the  latter  can  be  introduced  into  prob- 
lems involving  the  buoyant  effort  of  a  liquid  the  criterion  as 
to  whether  a  homogeneous  solid  will  sink  or  rise,  when  im- 
mersed in  the  standard  liquid,  is  more  easily  applied,  thus  : 
Being  immersed,  the  volume  V  of  the  body  =  that,  F,  of 
displaced  liquid.  Hence, 


590  MECHANICS   OF   ENGINEERING. 

if  G'  is  >  V'y,  i.e.,  if   V'y'  is  >  V'y,  or  a-  >  1,  it  sinks  ; 
while  if  G-'  is  <  V'y,    ......  or  tr  <  1,  it  rises  ; 

i.e.,  according  as  the  weight  G'  is  >  or  <  than  the  buoyant 
effort. 

Other  methods  of  determining  the  specific  gravity  of  solids, 
liquids,  and  gases  are  given  in  works  on  Physics. 

45"9.  Equilibrium  of  Flotation.  —  In  case  the  weight  G'  of  an 
immersed  solid  is  less  than  the  buoyant  effort  V'y  (where  V  is 
the  volume  of  displacement,  and  /the  heaviness  of  liquid)  the 
body  rises  to  the  surface,  and  after  a  series  of  oscillations  comes 
to  rest  in  such  a  position,  Fig.  503,  that  its  centre  of  gravity  C 
and  the  centre  of  buoyancy  B  (the  new  B,  belonging  to  the 
new  volume  of  displacement,  which  is  limited  above  by  the 
horizontal  plane  of  the  free  surface  of  the  liquid)  are  in  the 
same  vertical  (called  the  axis  of  flotation,  or  line  of  support), 
and  that  the  volume  of  displacement  has  diminished  to  such  a 
new  value  V,  that 

Vy=G'  ........    (1) 


In  the  figure,   V  —  vol.  AND,  below  the  horizontal  plane 
AN,  and  the  slightest  motion  of  the  body  will  change  the  form 
of  this  volume,  in  general  (whereas  with 
complete  immersion  the  volume  of  dis- 
placement remains  constant).     For  stable 
equilibrium  it  is  not  essential  in  every 
case  that  C  (centre  of  gravity  of  body) 
should  be  below  B  (the  centre  of  buoy- 
ancy) as  with  complete  immersion,  since  if 
FIG.  503.  "  the  solid  is  turned,  B  may  change  its  posi- 

tion in  the  body,  as  the  form  of  the  volume  AND  changes. 

There  is  now  no  definite  relation  between  the  volume  of 
displacement  Fand  that  of  the  body,  V,  unless  the  latter  is 
homogeneous,  and  then  for  G'  we  may  write  V'y',  i.e. 

V'y'  =  Vy  (for  a  homogeneous  solid)  ;     .     .     (2) 
or,  the  volumes  are  inversely  proportional  to  the  heavinesses. 


FLOTATION. 


591 


The  buoyant  effort  of  the  air  on  the  portion  ANE  may  be 
neglected  in  most  practical  cases,  as  being  insignificant. 

If  the  solid  is  hollow,  the  position  of  its  centre  of  gravity  C 
may  be  easily  varied  (by  shifting  ballast,  e.g.)  within  certain 
limits,  but  that  of  the  centre  of  buoyancy  J3  depends  only  on 
the  geometrical  form  of  the  volume  of  displacement  AND, 
below  the  horizontal  plane  AN. 

EXAMPLE.— (Ft.,  lb.,  sec.)  Will  a  solid  weighing  G  —  400 
Ibs.,  and  having  a  volume  V  =  8  cub.  feet,  without  hollows 
or  recesses,  float  in  water?  To  obtain  a  buoyant  effort  of 
400  Ibs.,  we  need  a  volume  of  displacement,  see  eq.  (1),  of 

G'       400 
^  =  —  =  ^  *  =  only  6.4  cub.  ft. 


!-* WATFR- 


Hence  the  solid  will  float  with  8  —  6.4,  or  1.6,  cub.  ft.  pro- 
jecting above  the  water  level. 

Query :  A  vessel  contains  water,  reaching  to  its  brim,  and 
also  a  piece  of  ice  which  floats  without  touching  the  vessel. 
When  the  ice  melts  will  the  water  overflow  2 

460.  The  Hydrometer  is  a  floating  instrument  for  determin- 
ing the  relative  heavinesses  of  liquids.  Fig.  504  shows  a  sim- 
ple form,  consisting  of  a  bulb  and  a  cylin- 
drical stem  of  glass,  so  designed  and 
weighted  as  to  float  upright  in  all  liquids 
whose  heavinesses  it  is  to  compare.  Let  F 
denote  the  uniform  sectional  area  of  the 
stem  (a  circle),  and  suppose  that  when  float- 
ing in  water  (whose  heaviness  =  y)  the 
water  surface  marks  a  point  A  on  the  stem  ; 
and  that  when  floating  in  another  liquid, 
say  petroleum,  whose  heaviness,  =  yp,  we 
wish  to  determine,  it  floats  at  a  greater 
depth,  the  liquid  surface  now  marking  A 
on  the  stem,  a  height  —  x  above  A.  G'  is 
the  same  in  both  experiments ;  but  while  the  volume  of  dis- 
placement in  water  is  F,  in  petroleum  it  is  V  -f-  Fx.  There- 
fore from  eq.  (1),  §  459, 


FIG.  504. 


592  MECHANICS   OF  ENGINEERING. 

in  the  water         G  —  Vy,        '.    ....     (1) 
and  in  the  petroleum  G  =  (  V+  Fx)yp  ;       .     .    (2) 

from  which,  knowing  G',  F,  x,  and  y,  we  find  Fand  yp,  i.e., 


(3) 


[N.B.  —  jPis  best  determined  by  noting  the  additional  dis- 
tance, =  I,  through  which  the  instrument  sinks  in  water  under 
an  additional  load  P,  not  immersed  ;  for  then 


EXAMPLE.  —  [Using  the  inch,  ounce,  and  second,  in  which 
system  y  =  1000  -=-  1728  =  0.578  (§  409).]  With  £'  =  3 
ounces,  and  F=  0.10  sq.  inch,  x  being  observed,  on  the 
graduated  stem,  to  be  5  inches,  we  have  for  the  petroleum 

3  X  0.578 


3  +  0.10X5X0.578 

=  56.7  Ibs.  per  cub.  foot. 

Temperature  influences  the  heaviness  of  most  liquids  to 
some  extent. 

In  another  kind  of  instrument  a  scale-pan  is  fixed  to  the  top 
of  the  stem,  and  the  specific  gravity  computed  from  the  weight 
necessary  to  be  placed  on  this  pan  to  cause  the  hydrometer  to 
sink  to  the  same  point  in  all  liquids  for  which  it  is  used. 

461,  Depth  of  Flotation.  —  If  the  weight  and  external  shape 
of  the  floating  body  are  known,  and  the  centre  of  gravity  so 
situated  that  the  position  of  flotation  is  known,  the  depth  of 
the  lowest  point  below  the  surface  may  be  determined. 


FLOTATION. 


593 


CASE  I.  Eight  prism  or  cylinder  with  its  axis  vertical. — 
Fig.  505.  (For  stability  in  this  position, 
see  §  464#.)  Let  G'  =  weight  of  cylin- 
der, j^the  area  of  its  cross-section  (full 
circle),  hf  its  altitude,  and  h  the  un- 
known depth  of  flotation  (or  draught) ; 
then  from  eq.  (1),  §  426, 


9L 

V 


in  which  y  —  heaviness  of  the  liquid. 
If    the  prism   (or  cylinder)  is  homo- 
geneous (and  then  (7,  at  the  middle  of  h ',  is  higher  than 
and  y  its  heaviness,  we  then  have 


FIG.  505. 


.    .    .    .    (2) 


in  which  cr  =  specific  gravity  of  solid  referred  to  the  liquid  as 
standard.     (See  §  458.) 

CASE  II.  Pyramid  or  cone  with  axis  vertical  and  vertex 
down. — Fig.  506.     Let  V  =  volume  of 
whole  pyramid  (or  cone),  and  V  =  vol- 
ume   of    displacement.     From    similar   ~~f" 
pyramids, 


/Tr 

But  O'  =  Vy\  or,  V=  — ;  whence 


FIG.  506. 


(3) 


594 


MECHANICS    OF    ENGINEERING. 


CASE  III.  Ditto,  ~but  vertex  up. — Fig.  50Y.  Let  the  nota- 
tion be  as  before,  for  V  and  V.  The 
part  out  of  water  is  a  pyramid  of  volume 
=  V"  =  V  —  V,  and  is  similar  to  the 
whole  pyramid  ; 


V-  Y:  Y'  ::  A"3  :  h'9. 


FIG.  507. 


===^     Also, 


.-.,  finally,  h  =  A'pL  -  j/1  -  [#'  -f-  F>]~|.    ...    (4) 

CASE  IY.  /Sphere. — Fig.  508.    The  volume  immersed  is 
Y  = 


and  hence,  since  Yy  =  Gf  =  weight 
of  sphere, 


„       nW       Gf 

Ttrh  —  =  — . 

3          y 


(5) 


From  which  cubic  equation  h  may  be 
^Ji*Hz:  obtained  by  successive  trials  and  ap- 


FlG-  m-  [An  exact  solution  of  (5)  for  the 

unknown  h  is  impossible,  as  it  falls  under  the  irreducible  case 
of  Cardan's  Rule.] 

CASE  Y.  Right  cylinder  with  axis  horizontal.  —  Fig.  509. 

Ie°r,  J 


of  seg.  Aj)B\  X 


Fia.509. 


hence,  since  F=  — , 

r 

-     sin  2a   =  — . 


EXAMPLES   OF  FLOTATION.  595 

Prom  this  transcendental  equation  we  can  obtain  <*,  by  trial, 
in  radians  (see  example  in  §  428),  and  finally  A,  since 

h  =  r(l  —  cos  a).  ' (7) 

EXAMPLE  1. — A  sphere  of  40  inches  diameter  is  observed  to 
have  a  depth  of  flotation  h  =  9  in.  in  water.  Required  its 
weight  G' .  From  eq.  (5)  (inch,  lb.,  sec.)  we  have 

G'  =  [62.5  +  1728]7r92[20  -  -J  X  9]  =  156.5  Ibs. 

The  sphere  may  be  hollow,  e.g.,  of  sheet  metal  loaded  with 
shot ;  constructed  in  any  way,  so  long  as  G'  and  the  volume 
T^of  displacement  remain  unchanged.  But  if  the  sphere  is 
homogeneous,  its  heaviness  (§  7)  y'  must  be 

=  G'  -f-  V  =  G1  -f-  f-Trr3  =  (156.5)  -f-  f  tf203 
=  .00466  Ibs.  per  cubic  inch, 

and  hence,  referred  to  water,  its  specific  gravity  is  <r  =  about 
0.13. 

EXAMPLE  2. — The  right  cylinder  in  Fig.  509  is  homogeneous 
and  10  inches  in  diameter,  and  has  a  specific  gravity  (referred 
to  water)  of  <r  =  0.30.  Required  the  depth  of  flotation  A. 

Its  heaviness  must  be  yf  =  cry ;  hence  its  weight 

G'  =  Very  =  nrHay ; 
hence,  from  eq,  (6), 

r*l\_a  —  f  sin  2#]  =  nr'lo',  .'.  OL  —  \  sin  2«  =  nv 

(involving  abstract  numbers  only).  Trying  a  —  60°  (  =  \tt  in 
radians),  we  have 

\n  —  %  sin  120°  =  0.614 ;  whereas  no-  =  .9424 

For  a  =  70°,  1.2217  -  i  sin  140°  =  0.9003  ; 
For  a  =  71°,  1.2391  -  i  sin  142°  =  0.9313  ; 
For  a  =  71°  22r,  1.2455  —  -J-  sin  142°  44r  =  0.9428,  which  may 
be  considered  sufficiently  close.     Now  from  eq.  (7), 

h  =  (5  in.)  (1  —  cos  71°  22')  =  3.40  iu.—Ans. 


596  MECHANICS   OF   ENGINEERING. 

462.  Draught  of  Ships. — In  designing  a  ship,  especially  if  of 
a  new  model,  the  position  of  the  centre  of  gravity  is  found  by 
eq.  (3)  of  §  23  (with  weights  instead  of  volumes) ;  i.e.,  the  sum 
of  the  products  obtained  by  multiplying  the  weight  of  each 
portion  of  the  hull  and  cargo  by  the  distance  of  its  centre  of 
gravity  from  a  convenient  reference-plane  (e.g.,  the  horizontal 
plane  of  the  keel  bottom)  is  divided  by  the  sum  of  the  weights, 
and  the  quotient  is  the  distance  of  the  centre  of  gravity  of  the 
whole  from  the  reference-plane. 

Similarly,  the  distance  from  another  reference-plane  is  de- 
termined. These  two  co-ordinates  and  the  fact  that  the  centre 
of  gravity  lies  in  the  median  vertical  plane  of  symmetry  of  the 
ship  (assuming  a  symmetrical  arrangement  of  the  framework 
and  cargo)  fix  its  location.  The  total  weight,  G',  equals,  of 
course,  the  sum  of  the  individual  weights  just  mentioned.  The 
centre  of  buoyancy,  for  any  assumed  draught  and  correspond- 
ing position  of  ship,  is  found  by  the  same  method;  but  more 
simply,  since  it  is  the  centre  of  gravity  of  the  imaginary  homo- 
geneous volume  between  the  water-line  plane  and  the  wetted 
surface  of  the  hull.  This  volume  (of  "displacement")  is 
divided  into  an  even  number  (say  4  to  8)  of  horizontal  laminae 
of  equal  thickness,  and  Simpson's  Rule  applied  to  find  the  vol- 
ume (i.e.,  the  V  of  preceding  formulae),  and  also  (eq.  3,  §  23) 
the  height  of  its  centre  of  gravity  above  the  keel.  Similarly, 
by  division  into  (from  8  to  20)  vertical  slices,  ~I  to  keel  (an 
even  number  and  of  equal  thickness),  we  find  the  distance  of 
the  centre  of  gravity  from  the  bow.  Thus  the  centre  of  buoy- 
ancy is  fixed,  and  the  corresponding  buoyant  effort  Vy  (tech- 
nically called  the  displacement  and  usually  expressed  in  tons) 
computed,  for  any  assumed  draught  of  ship  (upright).  That 
position  in  which  the  "  displacement"  =  G'  =  weight  of  ship 
is  the  position  of  equilibrium  of  the  ship  when  floating  up- 
right in  still  water,  and  the  corresponding  draught  is  noted. 
As  to  whether  this  equilibrium  is  stable  or  unstable,  the  fol- 
lowing will  show. 

In  most  ships  the  centre  of  gravity  C  is  several  feet  above 
the  centre  of  buoyancy,  B,  and  a  foot  or  more  below  the  water 
line. 


DRAUGHT   OF   SHIPS. 


597 


After  a  ship  is  afloat  and  its  draught  actually  noted  its  total 
weight  G',  =  Vy,  can  be  computed,  the  values  of  Ffor  dif- 
ferent draughts  having  been  calculated  in  advance.  In  this 
way  the  weights  of  different  cargoes  can  also  be  measured. 

EXAMPLE. — A  ship  having  a  displacement  of  5000  tons  is 
itself  5000  tons  in  weight,  and  displaces  a  volume  of  salt  water 
F=  #'-5-  y  =  10,000,000  Ibs.  H-  64  Ibs.  per  cub.  ft.  =  156250 
cub.  ft. 

463.  Angular  Stability  of  Ships.— If  a  vessel  floating  upright 
were  of  the  peculiar  form  and  position  of 
Fig.  510  (the  water-line  section  having  an 
area  =  zero)   its   tendency   to   regain   that 
position,  or  depart  from  it,  when  slightly 
inclined  an  angle  0  from  the  vertical  is  due 
to  the  action  of  the  couple  now  formed  by 
the  equal  and  parallel  forces  Vy  and  G' , 
which  are  no  longer  directly  opposed.     This 
couple  is  called  a  righting  couple  if  it  acts 
to  restore  the  first  position  (as  in  Fig.  511, 
where  C  is  lower  than  _#),  and  an 
upsetting    couple   if   the   reverse,  C 
above  B.     In    either  case  the  mo- 
ment of  the  couple  is 

=  Vy  .  BC  sin  0  •=.  Vye  sin  0,       t— —  — 

and  the  centre  of  buoyancy  B  does  not 
change  its  position  in  the  vessel,  since 
the  water-displacing  shape  remains 
the  same ;  i.e.,  no  new  portions  of 
the  vessel  are  either  immersed  or 
raised  out  of  the  water. 

But  in  a  vessel  of  ordinary  form,  when  turned  an  angle  0  from 
the  vertical,  Fig.  512  (in  which  ED  is  a  line  which  is  vertical 
when  the  ship  is  upright),  there  is  a  new  centre  of  buoyancy, 
Bl ,  corresponding  to  the  new  shape  A^NXD  of  the  displacement- 
volume,  and  the  couple  to  right  the  vessel  (or  the  reverse) 


FIG.  510. 


FIG.  511. 


598 


MECHANICS   OF   ENGINEERING. 


consists  of  the  two  forces  G'  at  C  and  Vy  at  JS, ,  and  has  a 

moment  (which  we  may  call  J/,  or 
/E  moment  of   stability)   of  a   value 

(§28) 


4=      M=Vy. 


FIG.  512. 


(1) 


^Lzrz  Now  conceive  put  in  at  B  (centre 
of  buoyancy  of  the  upright  posi- 
tion) two  vertical  and  opposite 
forces,  each  =  Vy  =  G\  calling 
them  P  and  Pl  (see  §  20),  Fig.  512. 
We  can  now  regard  the  couple  \_G\  Vy]  as  replaced  by  the 
two  couples  \G',  P]  and  [JP,,  Vy\\  for  evidently 

Vy  .  mC  sin   0  =  Vy  .  BC  sin  0  -f-  Vy  .  mJ3  sin  0  ; 
(§§33  and  34;) 

.-.  M  =  Vy  JtUsin  0  +  Vy^nB  sin  0.    .     .     (2) 

But  the  couple  [#',  P]  would  be  the  only  one  to  right  the 
vessel  if  no  new  portions  of  the  hull  entered  the  water  or 
emerged  from  it,  in  the  inclined  position;  hence  the  other 
couple  [JP15  Vy]  owes  its  existence  to  the  emersion  of  the 

wedge  AOA^  and  the  immersion 
of  the  wedge  NON^\  i.e.,  to  the 
loss  of  a  buoyant  force  Q  =  (vol- 
ume A  OAJ  X  y  on  one  side,  and  the 
—  gain  of  an  equal  buoyant  force  on 
the  other;  therefore  this  couple 
[jP,,  Vy]  is  the  equivalent  of  the 
couple  [Q,  Q],  Fig.  51L,  formed  by 
putting  in  at  the  centre  of  buoyancy 
of  each  of  the  two  wedges  a  vertical 
force 


Fio.  513. 


Q  =  (vol.  of  wedge)  X  r  =  Vwy.    (See  figure.) 


STABILITY   OF   SHIPS.  599 

If  a  denotes  the  arm  of  this  couple,  we  may  write 


Vy  .  mB  sin  0,  [of  eq.  (2)],  =  Vwya  ;     .     .     (3) 
and  hence,  denoting  .Z?(7by  e,  we  have 

M=  ±  Vyesiu  0  +  Vwya;    .    ..'.'.    (4) 

the  negative  sign  in  which  is  to  be  used  when  C  is  above  B 
(as  with  most  ships).  0,  the  intersection  of  ED  and  AN, 
does  not  necessarily  lie  on  the  new  water-line  plane  AlNl . 

EXAMPLE. — If  a  ship  of  ( Vy  =)  3000  tons  displacement 
with  C  4  ft.  above  B  (i.e.,  e  =  —  4  ft.)  is  deviated  10°  from 
the  vertical,  in  salt  water,  for  which  angle  the  wedges  A  OAl  and 
NON^  have  each  a  volume  of  4000  cubic  feet,  while  the  hori- 
zontal distance  a  between  their  centres  of  buoyancy  is  18  feet, 
the  moment  of  the  acting  couple  will  be,  from  eq.  (4)  (ft.-ton- 
sec.  system,  in  which  y  of  salt  water  =  0.032), 

M  =  —  3000  X  4  X  0.1736  +  4000  X  0.032  X 18  =  220.8  ft.  tons, 
which  being  -|-  indicates  a  righting  couple. 

464.  Remark. — If  with  a  given  ship  and  cargo  this  moment 
of  stability,  _3/,  be  computed,  by  eq.  (4),  for  a  number  of  values 
of  0,  and  the  results  plotted  as  ordinates  (to  scale)  of  a  curve, 
0  being  the  abscissa,  the  curve  ob- 
tained is  indicative  of  the  general 
stability  of  the  ship.  See  Fig.  514. 
For  some  value  of  0=  OK  (as  well 
as  for  0  =  0)  the  value  of  M  is 


zero,  and  for  0  >  OK,  M  is  nega-          ~^~"  -w\j 

tive,  indicating  an  upsetting  couple.  FlG- 514- 

That  is,  for  0=0  the  equilibrium  is  stable,  but  for  0  =  OK, 
unstable  ;  and  M  =  0  in  both  positions.  From  eq.  (4)  we  see 
why,  if  C  is  above  JE>,  instability  does  not  necessarily  follow. 

464a.  Metacentre  of  a  Ship. — Eeferring  again  to  Fig.  512, 
we  note  that  the  entire  couple  [#',  Vy]  will  be  a  righting 
couple,  or  an  upsetting  couple,  according  as  the  point  m  (the 


600  MECHANICS    OF    ENGINEERING. 

intersection  of  the  vertical  through  Bl  ,  the  new  centre  of 
buoyancy,  with  BC  prolonged)  is  above  or  below  the  centre 
of  gravity  C  of  the  ship.  The  location  of  this  point  m  changes 
with  0  ;  but  as  0  becomes  very  small  (and  ultimately  zero)  m 
approaches  a  definite  position  on  the  line  DE,  though  not  oc- 
cupying it  exactly  till  0  =  0.  This  limiting  position  of  m  is 
called  the  metacentre,  and  accordingly  the  following  may  be 
stated  :  A  ship  floating  iipright  is  in  stable  equilibrium  if  its 
metacentre  is  above  its  centre  of  gravity  ;  and  vice  versa. 
In  other  words,  for  a  slight  inclination  from  the  vertical  a 
righting,  and  not  an  upsetting,  couple  is  called  into  action,  if 
m  is  above  C.  To  find  the  metacentre,  by  means  of  the  dis- 
tance Bin,  we  have,  from  eq.  (3), 


-, 
Vy  sm  0 


(5) 


and  wish  ultimately  to  make  0  =  0.  Now  the  moment 
(  Vwy)db  =  the  sum  of  the  moments  about  the  horizontal  fore- 
and-aft  water-line  axis  OL,  Fig.  515,  of  the  buoyant  efforts 

due  to  the  immersion   of   the 
separate     vertical     elementary 
prisms  of  the  wedge  OLN^N, 
plus  the  moments  of  those  lost, 
from  emersion,  in  the   wedge 
.,.  ___      OLA.A.     Let  OA.LN,  be  the 
new  water-line  section  of  the 
ship    when    inclined    a   small 
FIG.  sis.  angle    0    from      the    vertical 

(0  =  NO,N^  and  OALN  the  old  water-line.  Let  2  =  the 
"1  distance  of  any  elementary  area  dF  oi  the  water-line  section 
from  OL  (which  is  the  intersection  of  the  two  water-line 
planes).  Each  dF  is  the  base  of  an  elementary  prism,  with 
altitude  =  (f>z,  of  the  wedge  N^OLN  (or  of  wedge  A^OLA 
when  z  is  negative).  The  buoyant  effort  of  this  prism  =  (its 
vol.)  X  y  =  yz<f>dF,  and  its  moment  about  OL  is 
Hence  the  total  moment,  =  Qa,  or  Vwya,  of  Fig.  505, 


=  70  X  I 


OL 


THE   METACENTRE.  601 

of  water-line  section,  in  which  IOL  denotes  the  "  moment  of 
inertia"  (§  85)  of  the  plane  figure  OALNO  about  the  axis  OL. 
Hence  from  (5),  putting  0  =  sin  0  (true  when  0  =  0),  we  have 
mB  =  IOL  -T-  V\  and  therefore  the  distance  mC\  of  the  meta- 
centre  m  above  (7,  the  centre  of  gravity  of  the  ship,  Fig.  512,  is 

~~n       -L          IOL  (of  water-line  sec.)    .  f~ 

mC,  =  Am,  =  Z°*±  --  _  --  1  ±  e,  ...     (6) 

in  which  e  =  BC  =  distance  from  the  centre  of  gravity  to  the 
centre  of  buoyancy,  the  negative  sign  being  used  when  C  is 
above  B\  while  V=  whole  volume  of  water  displaced  by  the 
ship. 
We  may  also  write,  from  eqs.  (6)  and  (1),  for  small  values 


Mom.  of  righting  couple  =  M  =  Vy  sin  0    -~  ±  e   ,  .    (7) 

or 

M=ysm  0[/0i±  Ve]  .....     (7)' 


Eqs.  (7)  and  (7)'  will  give  close  approximations  for  0  <  10°  or 
15°  with  ships  of  ordinary  forms. 

EXAMPLE   1.  —  A    homogeneous    right   parallelepiped,    of 
heaviness    yr,  floats    upright    as  in 
Fig.      516.        Find      the     distance 

mC  =  hm  for  its  metacentre  in  this     A  ^— \ M f\ 

position,  and  whether  the  equilibrium  -^ 

is  stable.     Here  the  centre  of  gravity,  = 

C,  being  the  centre  of  figure,  is  of  ; 

course  above  B,  the  centre  of  buoy-  i>z 

ancy  ;  hence  e  is  negative.     B  is  the   =^ 

centre  of  gravity  of  the  displacement, 

and  is  therefore  a  distance  ^h  below 

the  water-line.     We  here  assume  that  I  is  greater  than 

From  eq.  (2),  §  461, 


•B--"- 


•DJ, 


602  MECHANICS    OF   ENGINEERING. 

and  since  CD  =  %h',  and  BD  —  %h,  .',  e  =  i(A''—  A); 


.e.  «  = 


while  (§  90)  IOL,  of  the  water-line  section  AN,  =  -£% 
Also, 


and  hence,  from  eq.  (6),  we  have 


Hence 


if  V  is  >  6A/a  £  fl  —  £  V  the  position  in  Fig.  516  is 

r  v      7y 

one  of  stable  equilibrium,  and  vice  versa.     E.g.,  if  yf  =  Jy, 
#'  =  12  inches  and  A7  =  6  inches,  we  have  (inch,  pound,  sec.) 


hm  =  ^C  =  jf  [144  -  6  X 


=  2.5  in. 


The  equilibrium  will  be  unstable  if,  with  y'  •=.  %y,  b'  is  made 
less  than  1.225  ti  ';  for,  putting  mC  =  Q,  we  obtain  Jr  = 
1.225  A'. 

EXAMPLE  2.  —  (Ft.,  lb.,  sec.)     Let  Fig.  517  represent  the  half 
water-line  section  of  a  loaded  ship  of  G'  =  Vy  =  1010  tons 


FIG.  517. 


displacement;  required  the  height  of  the  metacentre  above  the 
centre  of  buoyancy,  i.e.,  mB  —  2  (See  equation  just  before  eq. 
(6).)  Now  the  quantity  /OL,  of  the  water-line  section,  may, 
from  symmetry,  (see  §  93,)  be  written 


IL  = 


(1) 


METACENTRE.  603 

in  which  y  =  the  ordinate  1  to  the  axis  OL  at  any  point;  and 
this,  again,  by  Simpson's  Rule  for  approximate  integration, 
OL  being  divided  into  an  even  number,  n,  of  equal  parts,  and 
ordinates  erected  (see  figure),  may  be  written 

/      -?.    fr£-° 
L       3  *       3n 


From  which,  by  numerical  substitution  (see  figure  for  dimen- 
sions ;  n  =  8), 


125 

1728  729 

or'                                   2197  2744 

343  1331 


IOL  =  W>.125  +4  X  4393  +  2  X  4804+  0.125] 


.       =  120801  biquad.  ft,,          =         = 

=  3.8  feet. 

That  is,  the  metacentre  is  3.8  feet  above  the  centre  of  buoyancy, 
and  hence,  if  J3C=2  feet,  is  1.90  ft.  above  the  centre  of 
gravity.  [See  Johnson's  Cyclopaedia,  article  Naval  Architec- 

ture.'} 

465.  Metacentre  for  Longitudinal  Stability.  —  If  we  consider 
the  stability  of  a  vessel  with  respect  to  pitching,  in  a  manner 
similar  to  that  just  pursued  for  rolling,  we  derive  the  position 
of  the  metacentre  for  pitching  or  for  longitudinal  stability  — 
and  this  of  course  occupies  a  much  higher  position  than  that 
for  rolling,  involving  as  it  does  the  moment  of  inertia  of  the 
water-line  section  about  a  horizontal  gravity  axis  ~|  to  the  keel. 
"With  this  one  change,  eq.  (6)  holds  for  this  case  also.  In 
large  ships  the  height  of  this  metacentre  above  the  centre  of 
gravity  of  the  ship  may  be  as  groat  as  90  feet. 


CHAPTER  Y. 

HYDROSTATICS  (Continued)— GASEOUS  FLUIDS. 

466.  Thermometers. — The  temperature,  or  "  liotness"  of 
liquids  has,  within  certain  limits,  but  little  influence  on  their 
statical  behavior,  but  with  gases  must  always  be  taken  into 
account,  since  the  three  quantities,  tension,  temperature,  and 
volume,  of  a  given  mass  of  gas  are  connected  by  a  nearly  in- 
variable law,  as  will  be  seen. 

An  air-thermometer,  Fig.  518,  consists  of  a  large  glass  bulb 
filled  with  air,  from  which  projects  a  fine  straight  tube  of 

even  bore  (so  that  equal  lengths 
represent  equal  volumes).  A 
small  drop  of  liquid,  A,  sepa- 
rates the  internal  from  the  ex- 
ternal air,  both  of  which  are 
at  a  tension  of  (say)  one  at- 
mosphere (14.7  Ibs.  per  sq.  inch).  When  the  bulb  is  placed 
in  melting  ice  (freezing-point)  the  drop  stands  at  some  point  F 
in  the  tube ;  when  in  boiling  water  (boiling  under  a  pressure 
of  one  atmosphere),  the  drop  is  found  at  B,  on  account  of  the 
expansion  of  the  internal  air  under  the  influence  of  the  heat 
imparted  to  it.  (The  glass  also  expands,  but  only  about  -3-^5- 
as  much ;  this  will  be  neglected.)  The  distance  FB  along  the 
tube  may  now  be  divided  into  a  convenient  number  of  equal 
parts  called  degrees.  If  into  one  hundred  degrees,  it  is  found 
that  each  degree  represents  a  volume  equal  to  the  -n^VW 
(.00367)  part  of  the  total  volume  occupied  by  the  air  at  freez- 
ing-point ;  i.e.,  the  increase  of  volume  from  the  temperature  of 
freezing-point  to  that  of  the  boiling-point  of  water  =  0.367  of  the 
volume  at  freezing,  the  pressure  being  the  same,  and  even  having 
any  value  whatever  (as  well  as  one  atmosphere),  within  ordi- 
nary limits,  so  long  as  it  is  the  same  both  at  freezing  and  boil- 

604 


THERMOMETERS.  605 

ing.  It  must  be  understood,  however,  that  by  temjjerature  of 
boiling  is  always  meant  that  of  water  boiling  under  one  at- 
mosphere pressure.  Another  way  of  stating  the  above,  if  one 
hundred  degrees  are  used  between  freezing  and  boiling,  is  as 
follows  :  That  for  each  degree  increase  of  temperature  the  in- 
crease of  volume  is  ^fj  of  the  total  volume  at  freezing ;  273 
being  the  reciprocal  of  .00367. 

As  it  is  not  always  practicable  to  preserve  the  pressure  con- 
stant under  all  circumstances  with  an  air-thermometer,  we  use 
the  common  mercurial  thermometer  for  most  practical  pur- 
poses. In  this,  the  tube  is  sealed  at  the  outer  extremity,  with 
a  vacuum  above  the  column  of  mercury,  and  its  indications 
agree  very  closely  with  those  of  the  air-thermometer.  That 
equal  absolute  increments  of  volume  should  imply  equal  incre- 
ments of  heat  imparted  to  these  thermometric  fluids  (under 
constant  pressure)  could  not  reasonably  be  asserted  without 
satisfactory  experimental  evidence.  This,  however,  is  not  al- 
together wanting,  so  that  we  are  enabled  to  say  that  within  a 
moderate  range  of  temperature  equal  increments  of  heat  pro- 
duce equal  increments  of  volume  in  a  given  mass  not  only  of 
atmospheric  air,  but  of  the  so-called  "perfect"  or  "permanent" 
gases,  oxygen,  nitrogen,  hydrogen,  etc.  (so  named  before  it  was 
found  that  they  could  be  liquefied).  This  is  nearly  true  for 
mercury  also,  and  for  alcohol,  but  not  for  water.  Alcohol  has 
never  been  frozen,  and  hence  is  used  instead  of  mercury  as  a 
thermometric  substance  to  measure  temperatures  below  the 
freezing-point  of  the  latter. 

The  scale  of  a  mercurial  thermometer  is  fixed  ;  but  with  an 
air-thermometer  we  should  have  to  use  a  new  scale,  and  in  a 
new  position  on  the  tube,  for  each  value  of  the  pressure. 

467,  Thermometric  Scales.— In  the  Fahrenheit  scale  the  tube 
between  freezing  and  boiling  is  marked  off  into  180  equal 
parts,  and  the  zero  placed  at  32  of  these  parts  below  the  freez- 
ing point,  which  is  hence  -f-  32°,  and  the  boiling-point  +  212°. 

The  Centigrade,  or  Celsius,  scale,  which  is  the  one  chiefly 
used  in  scientific  practice,  places  its  zero  at  freezing,  and  100° 
at  boiling-point.  Hence  to  reduce 


f306 


MECHANICS    OF   ENGINEERING. 


Fahr.  readings  to  Centigrade,  subtract  32°  and  multiply  by  -J ; 
Cent.  "  "  Fahrenheit,  multiply  by  £  and  add  32°. 

468.  Absolute  Temperature. — Experiment  also  shows  that  if 
a  mass  of  air  or  other  perfect  gas  is  confined  in  a  vessel  whose 
volume  is  but  slightly  affected  by  changes  of  temperature, 
equal  increments  of  temperature  (and  therefore  equal  incre- 
ments of  heat  imparted  to  the  gas,  according  to  the  preceding 
paragraph)  produce  equal  increments  of  tension  (i.e.,  pressure 
per  unit  area) ;  or,  as  to  the  amount  of  the  increase,  that  when 
the  temperature  is  raised  by  an  amount  1°  Centigrade,  the  ten- 
sion is  increased  ^-|-¥  of  its  value  at  freezing-point.  Hence, 
theoretically,  an  ideal  barometer  (containing  a  liquid  unaffected 
by  changes  of  temperature)  communicating  with  the  confined 
gas  (whose  volume  practically  remains  constant)  would  by 
its  indications  serve  as  a  thermometer, 
Fig.  519,  and  the  attached  scale  could  be 
graduated  accordingly.  Thus,  if  the  col- 
umn stood  at  A  when  the  temperature 
was  freezing,  A  would  be  marked  0°  on 
the  Centigrade  system,  and  the  degree 
spaces  above  and  below  A  would  each 
FIG.  519.  _  ^i_^  Q£  foe  height  AB,  and  therefore 

the  point  B  (cistern  level)  to  which  the  column  would  sink  if 
the  gas-tension  were  zero  would  be  marked  —  273°  Centi- 
grade. 

But  a  zero-pressure,  in  the  Kinetic  Theory  of  Gases  (§  408), 
signifies  that  the  gaseous  molecules,  no  longer  impinging 
against  the  vessel  walls  (so  that  the  press.  =  0),  have  become 
motionless;  and  this,  in  the  Mechanical  Theory  of  Heat,  or 
Thermodynamics,  implies  that  the  gas  is  totally  destitute  of  heat. 
Hence  this  ideal  temperature  of  —  273°  Centigrade,  or  —  460° 
Fahrenheit,  is  called  the  Absolute  Zero  of  Temperature,  and  by 
reckoning  temperatures  from  it  as  a  starting-point,  our  formulas 
will  be  rendered  much  more  simple  and  compact.  Tempera- 
ture so  reckoned  is  called  absolute  temperature,  and  will  be 
denoted  by  the  letter  T.  Hence  the  following  rules  for  re- 
duction : 


GASES  AND   VAPORS.  607 

Absol.  temp.  T  in  Cent,  degrees  =  Ordinary  Cent.  +  273° ; 
Absol.  temp.  T  in  Fahr.  degrees  =  Ordinary  Fahr.  -f-  460°. 
For  example,  for  20°  Cent.,  T  =  293°  Abs.  Cent. 

469.  Distinction  Between  Gases  and  Vapors. — All  known 
gases  can  be  converted  into  liquids  by  a  sufficient  reduction  of 
temperature  or  increase  of  pressure,  or  both ;  some,  however, 
with  great  difficulty,  such  as  atmospheric  air,  oxygen,  hydro- 
gen, nitrogen,  etc.,  these  having  been  but  recently  (1878)  re- 
duced to  the  liquid  form.  A  vapor  is  a  gas  near  the  point  of 
liquefaction,  and  does  not  show  that  regularity  of  behavior 
under  changes  of  temperature  and  pressure  characteristic  of  a 
gas  when  at  a  temperature  much  above  the  point  of  liquefac- 
tion. All  gases  treated  in  this  chapter  (except  steam)  are  sup- 
posed in  a  condition  far  removed  from  this  stage.  The  fol- 
lowing will  illustrate  the  properties  of  vapors.  See  Fig.  520. 
Let  a  quantity  of  liquid,  say  water,  be  intro-  THERM. 
duced  into  a  closed  space,  previously  vacuous, 
of  considerably  larger  volume  than  the  water, 
and  furnished  with  a  manometer  and  ther- 
mometer. Vapor  of  water  immediately  be- 
gins to  form  in  the  space  above  the  liquid,  and 
continues  to  do  so  until  its  pressure  attains  a 
definite  value  dependent  on  the  temperature, 
and  not  on  the  ratio  of  the  volume  of  the  vessel  and  the  origi- 
nal volume  of  water ;  e.g.,  if  the  temperature  is  70°  Fahren- 
heit, the  vapor  ceases  to  form  when  the  tension  reaches  a  value 
of  0.36  Ibs.  per  sq.  inch.  If  heat  be  gradually  applied  to  raise 
the  temperature,  more  vapor  will  form  (with  ebullition ;  i.e., 
from  the  body  of  the  liquid,  unless  the  heat  is  applied  very 
slowly),  but  the  tension  will  not  rise  above  a  fixed  value  for 
each  temperature  (independent  of  size  of  vessel)  so  long  as 
there  is  any  liquid  left.  Some  of  these  corresponding  values, 
for  water,  are  as  follows :  For  a 

Fahr.  temp.      =   70°     100°     150°     212°     220°     287°    300° 
Tension  (Ibs  )    =  Q  36    Q  93     3  6g     m     n  2     5 
persq.  in.)  j 

=  one  atm. 

At  any  such  stage  the  vapor  is  said  to  be  saturated. 


608  MECHANICS    OF   ENGINEERING. 

Finally,  at  some  temperature,  dependent  on  the  ratio  of  the 
original  volume  of  water  to  that  of  the  vessel,  all  of  the  water 
will  have  been  converted  into  vapor  (i.e.,  steam);  and  if  the 
temperature  be  still  further  increased,  the  tension  also  increases 
and  no  longer  depends  on  the  temperature  alone,  but  also  on 
the  heaviness  of  the  vapor  when  the  water  disappeared.  The 
vapor  is  now  said  to  be  superheated,  and  conforms  more  in  its 
properties  to  perfect  gases. 

470.  Critical  Temperature. — From  certain  experiments  there 
seems  to  be  reason  to  believe  that  at  a  certain  temperature, 
called  the  critical  temperature,  different  for  different  liquids, 
all  of  the  liquid  in  the  vessel  (if  any  remains,  and  supposing 
the  vessel  strong  enough  to  resist  the  pressure)  is  converted 
into  vapor,  whatever  be  the  size  of  the  vessel.  That  is,  above 
the  critical  temperature  the  substance  is  necessarily  gaseous, 
in  the  most  exclusive  sense,  incapable  of  liquefaction  by  pres- 
sure alone ;  while  below  this  temperature  it  is  a  vapor,  and  lique- 
faction will  begin  if,  by  compression  in  a  cylinder  and  conse- 
quent increase  of  pressure,  the  tension  can  be  raised  to  a  value 
corresponding,  for  a  state  of  saturation,  to  the  temperature 
(in  such  a  table  as  that  just  given  for  water).  For  example,  if 
vapor  of  water  at  220°  Fahrenheit  and  tension  of  10  Ibs.  per 
sq.  inch  (this  is  superheated  steam,  since  220°  is  higher  than 
the  temperature  which  for  saturation  corresponds  to  p  —  10 
Ibs.  per  sq.  inch)  is  compressed  slowly  (slowly,  to  avoid  change 
of  temperature)  till  the  tension  rises  to  17.2  Ibs.  per  sq.  in., 
which  (see  above  table)  is  the  pressure  of  saturation  for  a  tem- 
perature of  220°  Fahrenheit  for  water- vapor,  the  vapor  is  satu- 
rated, i.e.,  liquefaction  is  ready  to  begin,  and  during  any  fur- 
ther slow  reduction  of  volume  the  pressure  remains  constant 
and  some  of  the  vapor  is  liquefied. 

By  "  perfect  gases,"  or  gases  proper,  we  may  understand, 
therefore,  those  which  cannot  be  liquefied  by  pressure  unac- 
companied by  great  reduction  of  temperature;  i.e.,  whose 
"  critical  temperatures"  are  very  low.  The  critical  temperature 
of  NO2,  or  nitrous  oxide  gas,  is  between  —  11°  and  +  8°  Cen- 
tigrade, while  that  of  oxygen  is  said  to  be  at  —  118°  Centi- 


LAW   OF    CHARLES.  609 

grade.  [See  p.  471,  vol.  122  of  the  Journal  of  the  franklin 
Institute.  For  an  account  of  the  liquefaction  of  oxygen,  etc., 
see  the  same  periodical,  January  to  June,  1878.] 

471.  Law  of  Charles  (and  of  Gay  Lussac). — The  mode  of  gradu- 
ation of  the  air-thermometer  may  be  expressed  in  the  follow- 
ing formula,  which  holds  good  (for  practical  purposes)  within 
the  ordinary  limits  of  experiment  for  a  given  mass  of  any 

perfect  gas,  the  tension  remaining  constant : 

y=Y,+  0.00367  V0t  =  F0(l  +  .00367*) ;..(!) 

in  which  ~F0  denotes  the  volume  occupied  by  the  given  mass 
at  freezing-point  under  the  given  pressure,  Y  its  volume  at 
any  other  temperature  t  Centigrade  under  the  same  pressure. 
Now,  273  being  the  reciprocal  of  .00367,  we  may  write 

(273+Q.  F..  T  <  press.  ).(2) 

y°      273    '  'T;~ro   '         1  const,  f' 

(see  §  468 ;)  in  which  T0  =  the  absolute  temperature  of  freezing- 
point,  =  273°  absolute  Centigrade,  and  T  the  absolute  tem- 
perature corresponding  to  t  Centigrade.  Eq.  (2)  is  also  true 
when  T  and  T0  are  both  expressed  in  Fahrenheit  degrees  (from 
absolute  zero,  of  course).  Accordingly,  we  may  say  that,  the 
pressure  remaining  the  same,  the  volume  of  a  given  mass  of 
gas  varies  directly  as  the  absolute  temperature. 

Since  the  weight  of  the  given  mass  of  gas  is  invariable  at  a 
given  place  on  the  earth's  surface,  we  may 

always  use  the  equation   Yy  =  Y0y0 , (3) 

pressure  constant  or  not,  and  hence  (2)  may  be  rewritten 

^  =  —-...  (press,  const.) ;    .    (4) 
Y         ^o 

i.e.,  if  the  pressure  is  constant,  the  heaviness  (and  therefore 
the  specific  gravity}  varies  inversely  as  the  absolute  tempera- 
ture. 


610 


MECHANICS   OF   ENGINEERING. 


Experiment  also  shows  (§  468)  that  if  the  volume  [and  there- 
fore the  heaviness,  eq.  (3)]  remains  constant,  while  the  tem- 
perature varies,  the  tension  p  will  change  according  to  the 
following  relation,  in  which  pQ  =  the  tension  when  the  tem- 
perature is  freezing : 


(5) 


t  denoting  the  Centigrade  temperature. 
as  before,  we  have 


Hence  transforming, 


J>  _    T^       j  vol., 

'   ' 


and 


(  heav.,  const. 


(6) 


or,  the  volume  and  heaviness  remaining  constant,  the  tension 
of  a  given  mass  of  gas  varies  directly  as  the  absolute  tempera- 
ture. This  is  called  the  Law  of  Charles  (or  of  Gay  Lussac). 

472.  General  Formulae  for  any  Change  of  State  of  a  Perfect  Gas. 
— If  any  two  of  the  three  quantities,  viz.,  volume  (or  heavi- 
ness), tension,  and  temperature,  are  changed,  the  new  value  of 
the  third  is  determinate  from  those  of  the  other  two,  according 
to  a  relation  proved  as  follows  (remember- 
ing that  henceforth  the  absolute  temperature 
only  will  be  used,  T,  §  468) :  Fig.  521. 
At  A  a  certain  mass  of  gas  at  a  tension  of 
j?0,  one  atmosphere,  and  absolute  tempera- 
ture T0  (freezing),  occupies  a  volume  V0 . 
Let  it  now  be  heated  to  an  absolute  temp. 
=  T',  without  change  of  tension  (expanding 
behind  a  piston,  for  instance).  Its  volume  will  increase  to  a 
value  "F  which  from  (2)  of  §  471  will  satisfy  the  relation 


FIG.  521. 


Z  -r]L 

V  ~~  T  * 

v  o          •*• 


(7) 


(See  B  in  figure.) 

Let  it  now  be  heated  without  change  of  volume  to  an  abso- 
lute temperature  T  (C  in  figure).     Its  volume  is  still  V,  but 


LAWS    OF   PERFECT   GASES.  611 

the  tension  has  risen  to  a  value  p,  such  that,  on  comparing  B 
and  C  by  eq.  (6),  we  have 


Combining  (7)  and  (8),  we  obtain  for  any  state  in  which  the 
tension  is _p,  volume  V,  and  absolute  temperature  T,  in 

(General)  .     .     .    %-= -  =^7fr -  5     or  ^=-  =  a  constant ;  .     (9) 

or 

(General)  ....     -*^L  w  —  ^n    n , (10) 

which,  since 

(General).     .      Vy  =--  V,y,  =  Vmym  =  VnYn ,     .     .     .     (11) 

is  true  for  any  change  of  state,  we  may  also  write 

(General)  .     .     .   J.      -£=  =  -=%r, (12) 

or 

P™    —    P*  n  Q\ 

ymTm-ynTn- 

These  equations  (9)  to  (13),  inclusive,  hold  good  for  any  state 
of  a  mass  of  any  perfect  gas  (most  accurately  for  air).  The 
subscript  0  refers  to  the  state  of  one-atmosphere  tension  and 
freezing-point  temperature,  in  and  n  to  any  two  states  what- 
ever (within  practical  limits) ;  y  is  the  heaviness,  §§  7  and  409, 
and  77the  absolute  temperature,  §  468. 

If  j?,  T7",  and  T  oi  equation  (9)  be  treated  as  variables,  and 
laid  off  to  scale  as  co-ordinates  parallel  to  three  axes  in  space, 
respectively,  the  surface  so  formed  of  which  (9)  is  the  equation 
is  a  hyperbolic  paraboloid. 

473.  Examples. — EXAMPLE  1. — What  cubic  space  will  be 
occupied  by  2  Ibs.  of  hydrogen  gas  at  a  tension  of  two  atmos- 
pheres and  a  temperature  of  27°  Centigrade? 


612  MECHANICS   OF  ENGINEERING. 

With  the  inch-lb.-sec.  system  we  have  p0  =  14.7  lbs.  per  sq. 
inch,  y.  =  [.0056  -f-  1728]  Ibs.  per  cubic  inch,  and  T0  =  273° 
absolute  Centigrade,  when  the  gas  is  at  freezing-point  at  one 
atmosphere  (i.e.,  in  state  sub-zero).  In  the  state  mentioned  in 
the  problem,  we  havej?  =  2  X  14.7  Ibs.  per  sq.  in., 

T  —  273  +  27  =  300°  absolute  Centigrade, 
while  y  is  required.     Hence,  from  eq.  (12), 

2  X  14.7  = 14.7 

Y  300     ~  (.0056  -r-  1728)273  ' 

/.  Y  =  '          Ibs.  per  cub.  in.  =  .0102  Ibs.  per  cub.  foot ;  and  if 

17-28 

the  total  weight,  =  6r,  =  Vy,  is  to  be  2  Ibs.,  we  have  (ft.,  lb., 
sec.)  V—  2  -f-.0102  =  196  cubic  feet.— Ans. 

EXAMPLE  2. — A  mass  of  air  originally  at  24°  Centigrade 
and  a  tension  indicated  by  a  barometric  column  of  40  inches 
of  mercury  has  been  simultaneously  reduced  to  half  its 
former  volume  and  heated  to  100°  Centigrade ;  required  its 
tension  in  this  new  state,  which  we  call  the  state  n,  in  being  the 
original  state.  Use  the  inch,  lb.,  sec.  "We  have  given,  there- 
fore, pm  =  nx  14-7  lbs-  Per  sq-  incll>  T™  =  2?3  +  24  =  29?° 

absolute  Centigrade,  the  ratio 

Vm  :  Vn  =  2  :  1,  and  Tn  =  273  +  100  =373°  Abs.  Cent.; 

while  pn  is  the  unknown  quantity.     From  eq.  (10),  hence, 

pn  =  ^ .  ^-  .pm  =  2  X  m  •  f  *  X  14.7  =  49.22  Ibs.  per  sq.  in., 

*  n      -*-m 

which  an  ordinary  steam-gauge  would  indicate  as 
(49.22  -  14.7)  =  34.52  lbs.  per  sq.  inch. 

(That  is,  if  the  weather  barometer  indicated  exactly  14.7  lbs. 
per  sq.  inch.) 


EXAMPLES.      PERFECT   GASES. 


613 


EXAMPLE  3. — A  mass  of  air,  Fig.  522,  occupies  a  rigid  closed 
vessel  at  a  temperature  of  15°  Centigrade  (equal  to  that  of  sur- 
rounding objects)  and  a  tension 
of  four  atmospheres  [state  m~\. 
By  opening  a  stop-cock  a  few 
seconds,  thus  allowing  a  portion 
of  the  gas  to  escape  quickly,  and  0 
then  shutting  it,  the  remainder  FIG.  522. 

of  the  air  [now  in  state  ri\  is  found  to  have  a  tension  of  only 
2.5  atmospheres  (measured  immediately) ;  its  temperature  can- 
not be  measured  immediately  (so  much  time  being  necessary 
to  affect  a  thermometer),  and  is  less  than  before.  To  compute 
this  temperature,  Tn,  we  allow  the  air  now  in  the  vessel  to 
come  again  to  the  same  temperature  as  surrounding  objects 
(15°  Centigrade) ;  find  then  the  tension  to  be  2.92  atmospheres. 
Call  the  last  state,  state  r  (inch,  lb.,  sec.).  The  problem  then 
stands  thus : 


pm  =  4  x  14.7 

rm  =  ? 

Tm  =  288°  Abs.  Cent. 


pn  =  2.5  X  14.7 

princip 
unknown 


n  _  y  j  principal 


Pr  =  2.92  X  14.7 

yr  =  yn  (since  Vr  =  Vn) 

Tr  =  Tm  =  288°  Abs.  Cent. 


In  states  n  and  r  the  heaviness  is  the  same ;  hence  an  equa* 
tion  like  (6)  of  §  471  is  applicable,  whence 


or  —  27°  Centigrade ;  considerably  below  freezing,  as  a  result  of 
allowing  the  sudden  escape  of  a  portion  of  the  air,  and  the  con- 
sequent sudden  expansion,  and  reduction  of  tension,  of  the  re- 
mainder. In  this  sudden  passage  from  state  m  to  state  n,  the 
remainder  altered  its  heaviness  (and  its  volume  in  inverse  ratio) 
in  the  ratio  (see  eqs.  (11)  and  (10)  of  §  472) 


•*  m 

~Tn 


2.5  X  14.7   288 
4  X  14.7  *  246 


ISTow  the  heaviness  in  state  m  (see  eq.  (12),  §  472)  was 


614 


MECHANICS   OF  ENGINEERING. 


V     - 

7m  — 


14.7    .0807    273 


.306 


•p. 


288 


1728    14.7       1728 


Ibs.  per  cub.  in.  =  .306  Ibs.  per  cub.  ft. 

.-.  yn  =  0.73  X  ym  =  0.223  Ibs.  per  cub.  ft., 


and  also,  since  Vm  =  0.73  T^,  about  T2T7F  of  the  original  quan- 
tity of  air  in  vessel  has  escaped. 

[JSToxE.  —  By  numerous  experiments  like  this,  the  law  of 
cooling,  when  a  mass  of  gas  is  allowed  to  expand  suddenly  (as, 
e.g.,  behind  a  piston,  doing  work)  has  been  determined  ;  and 
vice  versa,  the  law  of  heating  under  sudden  compression  ;  see 


T' 


it] 


aL 


474.  The  Closed  Air-manometer.  —  If  a  manometer  be  formed 
of  a  straight  tube  of  glass,  of  uniform  cylindrical  bore,  which 
is  partially  filled  with  mercury  and  then  inverted  in  a  cistern 
of  mercury,  a  quantity  of  air  having  been  left  between  the 

mercury  and  the  upper  end  of  the 
tube,  which  is  closed,  the  tension  of 
this  confined  air  (to  be  computed 
from  its  observed  volume  and  tem- 
perature) must  be  added  to  that  due 
to  the  mercury  column,  in  order  to 
obtain  the  tension^'  to  be  measured. 
See  Fig.  523.  The  advantage  of  this 
kind  of  instrument  is,  that  to  meas- 
ure great  tensions  the  tube  need  not 
be  very  long.  Let  the  temperature 
T7,  of  whole  instrument,  and  the  tension  p1  of  the  air  or  gas 
in  the  cistern,  be  known  when  the  mercury  in  the  tube  stands 
at  the  same  level  as  that  in  the  cistern.  The  tension  of  the 
air  in  the  tube  must  now  be  p,  also,  its  temperature  T7,  ,  and  its. 
volume  is  V1  =  Fh^  ,  F  being  the  sectional  area  of  the  bore  of 
the  tube  ;  see  on  left  of  figure.  "When  the  instrument  is  used, 
gas  of  unknown  tension  p'  is  admitted  to  the  cistern,  the  tem- 
perature of  the  whole  instrument  being  noted  (=  T\  and  the 
heights  h  and  h"  are  observed  (h  -f-  h"  cannot  be  put  =  //., 


FIG.  523. 


CLOSED   AIR-MANOMETER.  615 

unless  the  cistern  is  very  large),    p'  is  then  computed  as  fol- 
lows (eq.  (2),  §  413)  : 


(i) 


in  which  p  =  the  tension  of  the  air  in  the  tube,  and  ym  the 
heaviness   of   mercury.     But  from   eq.  (10),  §  472,  putting 


7,     T      h,  T 
f^T'-T^-ST 

Hence  finally,  from  (1)  and  (2), 


(3) 


Since  T7,  ,  pl  ,  and  hl  are  fixed  constants  for  each  instrument, 
we  may,  from  (3),  compute  p1  for  any  observed  values  of  h  and 
T(N".B.  T  and  Tt  are  absolute  temperatures),  and  construct 
a  series  of  tables  each  of  which  shall  give  values  of  p'  for  a 
range  of  values  of  A,  and  one  special  value  of  T. 

EXAMPLE.  —  Supposing  the  fixed  constants  of  a  closed  air- 
manometer  to  be  (in  inch-lb.-sec.  system)  p^  —  14.7  (or  one 
atmosphere),  T,  =  285°  Abs.  Cent,  (i.e.,  12°  Centigrade),  and 
Aj  —  3'  4"  =  40  inches  ;  required  the  tension  in  the  cistern 
indicated  by  h"  —  25  inches  and  k  —  15  inches,  when  the 
temperature  is  —  3°  Centigrade,  or  T  =  270°  Abs.  Cent. 

For  mercury,  ym  =  [848.7  -=-  1728]  (§  409)  (though  strictly 
it  should  be  specially  computed  for  the  temperature,  since  it 
varies  about  .00002  of  itself  for  each  Centigrade  degree). 
Hence,  eq.  (3), 


Ibs.  per  sq.  inch,  or  nearly  3J  atmospheres  [steam-gauge  would 
read  34.7  Ibs.  per  sq.  in.]. 

475.  Marietta's  Law,  (or  Boyle's,)  Temperature  Constant  ;  i.e., 
Isothermal  Change.  —  If   a  mass  of  gas  be  compressed,  or  al- 


616  MECHANICS   OF   ENGINEERING. 

lowed  to  expand,  isothermally,  i.e.,  without  change  of  tern, 
perature  (practically  this  cannot  be  done  unless  the  walls  of  the 
vessel  are  conductors  of  heat,  and  then  the  motion  must  be 
slow),  eq.  (10)  of  §  472  now  becomes  (since  Tm  =  Tn) 

(  MariottJs    Law,  )        TT  „      -  v  <n      nr^  —  Y^  •     n\ 
I      Temp,  constant  f  VnPn>        pn "  Vm ' 

i.e.,  the  temperature  remaining  unchanged,  the  tensions  are 
inversely  proportional  to  the  volumes,  of  a  given  mass  of  a 
perfect  gas  ;  or,  the  product  of  volume  by  tension  is  a  constant 
quantity.  Again,  since  Vmym  —  Vnyn  for  any  change  of 
state, 

(  .MariottJs    Law, )  Pm_Ym     Qr   Pm  __  Pn  .  /2\ 

[      Temp,  constant  j  pn  ~    yn '  ym      yn' 

i.e.,  the  pressures  (or  tensions  are  directly  proportional  to  the 
(first  power  of  the)  heavinesses,  if  the  temperature  is  the  same. 
This  law,  which  is  very  closely  followed  by  all  the  perfect 
gases,  was  discovered  by  Boyle  in  England  and  Mariotte  in 
France  more  than  two  hundred  years  ago,  but  of  course  is  only 
a  particular  case  of  the  general  formula,  for  any  change  of 
state,  in  §  472.     It  may  be  verified  experimen- 
p'   '  tally  in  several  ways.     E.g.,  in  Fig.  524,  the 

tube  0 M  being  closed  at  the  top,  while  PN  is 
open,  let  mercury  be  poured  in  at  P  until  it 
reaches  the  level  A '  B '.     The  air  in  OA  is  now 
at  a  tension  of  one  atmosphere.     Let  more  mer- 
cury be  slowly  poured  in  at  P,  until  the  air 
confined  in  0  has  been  compressed  to  a  volume 
OA"  =  i  of  OA,  and  the  height  B"E"  then 
measured  ;  it  will  be  found  to  be  30  inches ;  i.e., 
the  tension  of  the  air  in  O  is  now  two  atmos- 
pheres (corresponding  to  60  inches  of  mercury). 
FIG.  524.         Again,  compress  the  air  in  O  to  \  its  original 
volume  (when  at   one  atmosphere),  i.e.,  to  volume   OA'"  = 
\OA',  and  the  mercury  height  B'"E"'  will  be  60  inches,  show- 
ing a  tension  of  three  atmospheres  in  the  confined  air  at  0  (90 


MAKIOTTES  LAW. 


617 


inches  of  mercury  in  a  barometer).  It  is  understood  that  the 
temperature  is  the  same,  i.e.,  that  time  is  given  the  compressed 
air  to  acquire  the  temperature  of  surrounding  objects  after 
being  heated  by  the  compression,  if  sudden. 

[NOTE.  —  The  law  of  decrease  of  steam-pressure  in  a  steam- 
engine  cylinder,  after  the  piston  has  passed  the  point  of  "  cut- 
off "  and  the  confined  steam  is  expanding,  does  not  materially 
differ  from  Mariotte's  law,  which  is  often  applied  to  the  case 
of  expanding  steam  ;  see  §  479.] 

While  Mariotte's  law  may  be  considered  exact  for  practical 
purposes,  it  is  only  approximately  true,  the  amount  of  the 
deviations  being  different  at  different  temperatures.  Thus, 
for  decreasing  temperatures  the  product  Vp  of  volume  by 
tension  becomes  smaller,  with  most  gases. 

EXAMPLE  1.  —  If  a  mass  of  compressed  air  expands  in  a 
cylinder  behind  a  piston,  having  a  tension  of  60  Ibs.  per  sq. 
inch  (45.3  by  steam-gauge)  at  the  beginning  of  the  expansion, 
which  is  supposed  slow  (that  the  temperature  may  not  fall)  ; 
then  when  it  has  doubled  in  volume  its  tension  will  be  only 
30  Ibs.  per  sq.  inch  ;  when  it  has  tripled  in  volume  its  tension 
will  be  only  20  Ibs.  per  sq.  inch,  and  so  on. 

EXAMPLE  2.  Diving-bell.  —  Fig.  525.  If  the  cylindrical 
diving-bell  AB  is  10  ft.  in  height,  in  what 
depth,  h  =  ?,  of  salt  water,  can  it  be  let  down 
to  the  bottom,  without  allowing  the  water  to 
rise  in  the  bell  more  than  a  distance  a  =  4  ft.  ? 
Call  the  horizontal  sectional  area,  F.  The 
mass  of  air  in  the  bell  is  constant,  at  a  constant 
temperature.  First,  algebraically  ;  at  the 
surface  this  mass  of  air  occupied  a  volume 
Vm  =  FTi"  at  a  tension  pm  =  14.  Y  X  144  Ibs. 
per  sq.  ft.,  while  at  the  depth  mentioned  it  is 
compressed  to  a  volume  Vn  =  F(h"  —  a\  and 
is  at  a  tension  pn  =pm  +  (h  —  a)yw  ,  in  which 
yw  =  heaviness  of  salt  water.  Hence,  from 
eq.  (1), 


XT-- 


Vm- 


-d);    .     .    (3) 


618  MECHANICS   OF   ENGINEERING. 


hence,  numerically,  (ft..  lb.,  sec.,) 


476.  Mixture  of  Gases.  —  It  is  sometimes  stated  that  if  a  vessel 
is  occupied  by  a  mixture  of  gases  (between  which  there  is  no 
chemical  action),  the  tension  of  the  mixture  is  equal  to  the  sum 
of  the  pressures  of  each  of  the  component  gases  present  ;  or, 
more  definitely,  is  equal  to  the  sum  of  the  pressures  which  the 
separate  masses  of  gas  would  exert  on  the  vessel  if  each  in  turn 
occupied  it  alone  at  the  same  temperature. 

This  is  a  direct  consequence  of  Mariotte's  law,  and  may  be 
demonstrated  as  follows  : 

Let  the  actual  tension  be  p,  and  the  capacity  of  the  vessel  V. 
Also  let  Fj  ,  F2  ,  etc.,  be  the  volumes  actually  occupied  by  the 
separate  masses  of  gas,  so  that 

Vt  +  V,+  .  .  .  =  V;      .....    (1) 

and  Pupt,  etc.,  the  pressures  they  would  individually  exert 
when  occupying  the  volume  V  alone  at  the  same  tempera- 
ture. Then,  by  Mariotte's  law, 

Vpt=VlP-,     Vpt=V#;    etc.;     ...    (2) 
whence,  by  addition,  we  have 

...)  =  (Vi  +  F2+...f; 
i.e.,    1>  =?,+!>,  -Jr.--..     .     ....     (3) 


Of  course,  the  same  statement  applies  to  any  number  of 
separate  parts  into  which  we  may  imagine  a  mass  of  homo- 
geneous gas  to  be  divided. 

For  numerical  examples  and  practical  questions  in  the  solu- 
tion of  which  this  principle  is  useful,  see  p.  239,  etc.,  Ran- 
kine's  Steam-engine.  (Rankine  uses  0.365,  where  0.367  has 
been  used  here.) 


BAROMETRIC    LEVELLING.  619 

477.  Barometric  Levelling. — By  measuring  with  a  barometer 
the  tension  of  the  atmosphere  at  two  different  levels,  simul- 
taneously, and  on  a  still  day,  the  two  localities  not  being  widely 
separated  horizontally,  we  may  compute  their  vertical  distance 
apart  if  the  temperature  of  the  stratum  of  air  between  them 
is  known,  being  the  same,  or  nearly  so,  at  both  m 

stations.     Since  the  heaviness  of   the  air  is 
different  in  different   layers   of  the  vertical    : 
column  between  the  two  elevations  J^and  M,   ': 
Fig.  526,  we  cannot  immediately  regard  the     • 
whole  of  such  a  column  as  a  free  body  (as  was    ..  •  ILr:y;— <P+#P 
done  with  a  liquid,  §  412),  but  must  consider   '' 
a   horizontal   thin    lamina,   Z,    of   thickness    .'  '  " 


—  dz  and  at  a  distance  =z  (variable)  below 
Jf,  the  level  of  the  upper  station,  N  being 
the  lower  level  at  a  distance,  A,  from  M. 

The  tension,  p,  must  increase  from  M 
downwards,  since  the  lower  laminae  have  to  support  a  greater 
weight  than  the  upper ;  and  the  heaviness  y  must  also  increase, 
proportionally  to  JP,  since  we  assume  that  all  parts  of  the  col- 
umn are  at  the  same  temperature,  thus  being  able  to  apply 
Mariotte's  law.  Let  the  tension  and  heaviness  of  the  air  at 
the  upper  base  of  the  elementary  lamina,  Z,  be  p  and  y  re- 
spectively. At  the  lower  base,  a  distance  dz  below  the  upper, 
the  tension  is^>  -f-  dp.  Let  the  area  of  the  base  of  lamina  be 
F;  then  the  vertical  forces  acting  on  the  lamina  are  Fp,  down- 
ward ;  its  weight  yFdz  downward  ;  and  F(p  +  dp)  upward. 
For  its  equilibrium  ^(vert.  compons.)  must  =  0 ; 

/.  F(p  +  dp)  -  Fp  -  Fydz  =  0; 

i.e.,  dp  =  ydz, (1) 

which  contains  three  variables.     But  from  Mariotte's  law, 
§  475,  eq.  (2),  if  pn  and  yn  refer  to  the   air  at  N,  we  may 

substitute  y=^p  and  obtain,  after  dividing  by  p,  to  separate 

Pn 

the  variables/*  and  2, 


620  MECHANICS    OF   ENGINEERING. 

»:*«;*.  (2) 

Yn       P 

Summing  equations  like  (2),  one  for  each  lamina  between 
M  (where  p  =pm  and  s  =  0)  and  N(  where  p  =j}n&ndz  =  h\ 
we  have 


=  /V 


which  gives  A,  the  difference  of  level,  or  altitude,  between  M 
and  N,  in  terms  of  the  observed  tensions  pn  and^?TO,  and  of  yn  , 
the  heaviness  of  the  air  at  ^V,  which  may  be  computed  from 
eq.  (12),  §  472,  substituting  from  which  we  have  finally 


in  which  the  subscript  0  refers  to  freezing-point  and  one  at- 
mosphere tension  ;  Tn  and  T0  are  absolute  temperatures.  For 
the  ratio  pn  :  pm  we  may  put  the  equal  ratio  hn  :  hm  of  the 
actual  barometric  heights  which  measure  the  tensions.  The 
log.  e(or  Naperian,  or  natural,  or  hyperbolic,  log.)  —  (common 
log.  to  base  10)  X  2.30258.  From  §  394,  y0  of  air  —  0.0807ft 
Ibs.  per  cub.  ft.,  and^0  =  14.701  Ibs.  per  sq.  inch  ;  T0  =  273° 
Abs.  Cent. 

If  the  temperature  of  the  two  stations  (both  in  the  shade) 
are  not  equal,  a  mean  temp.  =  %(Tm-{-  Tn)  may  be  used  for 
Tn  in  eq.  (4),  for  approximate  results.  Eq.  (4)  may  then  be 
written 

...     (5) 
The  quantity  ^  =  26213  ft.,  just   substituted,  is   called   the 

Yo 

height  of  the  homogeneous  atmosphere,  i.e.,  the  ideal  height 
which  the  atmosphere  would  have,  if  incompressible  and  non- 


BAROMETRIC   LEVELLING  —  ADIABATIC   LAW.  621 

expansive  like  a  liquid,  in  order  to  exert  a  pressure  of  14.701 
Ibs.  per  sq.  inch  upon  its  base,  being  throughout  of  a  constant 
heaviness  =  .08076  Ibs.  per  cub.  foot. 
By  inversion  of  eq.  (4)  we  may  also  write 


.r. 

'rj 


(6) 


where  e  =  2.71828  =  the  Naperian  Base,  which  is  to  be  raised 

rri 

to  the  power  whose  index  is  the  abstract  number  —  .  —  -  .  A 

Po       Tn 

a,nd  the  result  multiplied  by  j?OT  to  obtain  pn. 

EXAMPLE.  —  Having  observed  as  follows  (simultaneously)  : 

At  lower  station  jV,  hn  =  30.05  in.  mercury  ;  temp.  =  77.6°  F.  ; 
"upper      «     M,hm  =  23.26"          "'         «      =  70.4°F.; 

required  the  altitude  h.  From  these  figures  we  have  a  mean 
absolute  temperature  of  460°  +  £(77.6  +  70.4)  =  534°  Abs. 
Fahr.  ;  hence,  from  (5), 


h  =  26213  X  Hi  X  2.30258  X  log.  10     —    =  6787.9  ft. 


(Mt.  Guanaxuato,  in  Mexico,  by  Baron  von  Humboldt.) 
Strictly,  we  should  take  into  account  the  latitude  of  the  place, 
since  y0  varies  with  g  (see  §  76),  and  also  the  decrease  in  the 
intensity  of  gravitation  as  we  proceed  farther  from  the  earth's 
centre,  for  the  mercury  in  the  barometer  weighs  less  per  cubic 
inch  at  the  upper  station  than  at  the  lower. 

Tables  for  use  in  barometric  levelling  can  be  found  in  Traut- 
wine's  Pocket-book,  and  in  Searles's  Field-book  for  Railroad 
Engineers,  as  also  tables  of  boiling-points  of  water  under  dif- 
ferent atmospheric  pressures,  forming  the  basis  of  another 
method  of  determining  heights. 

478.  Adiabatic  Change—  Poisson's  Law.—  By  an  adiabatic 
change  of  state,  on  the  part  of  a  gas,  is  meant  a  compression 
or  expansion  in  which  work  is  done  upon  the  gas  (in  compress- 


622  MECHANICS   OF   ENGINEERING. 

ing  it)  or  by  the  gas  (in  expanding  against  a  resistance)  when 
there  is  no  transmission  of  heat  between  the  gas  and  enclosing 
vessel,  or  surrounding  objects,  by  conduction  or  radiation. 
This  occurs  when  the  volume  changes  in  a  vessel  of  non-con- 
ducting material,  or  when  the  compression  or  expansion  takes 
place  so  quickly  that  there  is  no  time  for  transmission  of  heat 
to  or  from  the  gas. 

The  experimental  facts  are,  that  if  a  mass  of  gas  in  a  cylinder 
be  suddenly  compressed  to  a  smaller  volume  its  temperature  is 
raised,  and  its  tension  increased  more  than  the  change  of  vol- 
ume would  call  for  by  Mariotte's  law ;  and  vice  versa,  if  a  gas 
at  high  tension  is  allowed  to  expand  in  a  cylinder  and  drive  a 
piston  against  a  resistance,  its  temperature  falls,  and  its  tension 
diminishes  more  rapidly  than  by  Mariotte's  law. 

Again  (see  Example  3,  §  473),  if  -$fc  of  the  gas  in  a  rigid 
vessel,  originally  at  4  atmos.  tension  and  temperature  of 
15°  Cent.,  is  allowed  to  escape  suddenly  through  a  stop-cock 
into  the  outer  air,  the  remainder,  while  increasing  its  volume 
in  the  ratio  100  :  73,  is  found  to  have  cooled  to  —  27°  Cent., 
and  its  tension  to  have  fallen  to  2.5  atmospheres;  whereas,  by 
Mariotte's  law,  if  the  temperature  had  been  kept  at  288°  Abs. 
Cent.,  the  tension  would  have  been  lowered  to  y7^  of  4,  i.e., 
to  2.92  atmospheres  only. 

The  reason  for  this  cooling  during  sudden  expansion  is,  ac- 
cording to  the  Kinetic  Theory  of  Gases,  that  since  the  "  sensi- 
ble heat"  (i.e.,  that  perceived  by  the  thermometer),  or  "  hot- 
ness"  of  a,  gas  depends  on  the  velocity  of  its  incessantly  moving 
molecules,  and  that  each  molecule  after  impact  with  a  receding 
piston  has  a  less  velocity  than  before,  the  temperature  neces- 
sarily falls;  and  vice  versa,  when  an  advancing  piston  com- 
presses the  gas  into  a  smaller  volume. 

If,  however,  a  mass  of  gas  expands  without  doing  work,  as 
when,  in  a  vessel  of  two  chambers,  one  a  vacuum,  the  other 
full  of  gas,  communication  is  opened  between  them,  and  the 
gas  allowed  to  fill  both  chambers,  no  cooling  is  noted  in  the 
mass  as  a  whole  (though  parts  may  have  been  cooled  tem- 
porarily). 

By  experiments  similar  to  that  in  Example  3,  §  473,  it  has 


ADIABATIC   CHANGE  —  EXAMPLES.  623 

been  found  that  for  air  and  the  "  perfect  gases,"  in  an  adiabatic 
change  of  volume  [and  therefore  of  heaviness],  the  tension 
varies  inversely  with  the  1.41th  power  of  the  volume.  This. 
is  called  Poissorfs  Law.  For  ordinary  purposes  (as  Weisbach 
suggests)  we  may  use  f  instead  of  1.41,  and  hence  write 

Adiabat.  \  Pm  _  f7m\*  _  pm  _  (  Vn  \* 

Change]  £  ~  (*/'    '  T»  =  W'    '     ' 

and  combining  this  relation  with  the  general  eqs.  (10)  and  (13), 
§  472,  we  have  also 

Adiabat.  )  Pm  _  I  T^ 

Change  ]  '  pn  "  \~Tn)  > 

i.e.,  the  tension  varies  directly  as  the  cube  of  the  absolute  tem- 
perature; also, 

Adiabat.  )         /  KA  _  (Tn  \a          yn  _  iTn  \\ 
Change]         fe$^W'         £      \Ej' 

i.e.,  the  volume  is  inversely,  and  the  heaviness  directly,  as  the 
square  of  the  absolute  temperature. 

Here  m  and  n  refer  to  any  two  adiabatically  related  states. 
T  is  the  absolute  temperature. 

EXAMPLE  1.  —  Air  in  a  cylinder  at  20°  Cent,  is  suddenly 
compressed  to  -J-  its  original  volume  (and  therefore  is  six  times 
as  dense,  i.e.,  has  six  times  the  heaviness,  as  before).  To  what 
temperature  is  it  heated  ?  Let  m  be  the  initial  state,  and  n  the 
final.  From  eq.  (3)  we  have 


=  718°  Abe.  Cent, 


or  nearly  double  the  absolute  temperature  of  boiling  water. 

EXAMPLE  2.  —  After  the  air  in  Example  1  has  been  given 
time  to  cool  again  to  20°  Cent,  (temperature  of  surrounding 
objects)  it  is  allowed  to  resume,  suddenly,  its  first  volume,  i.e., 


624  MECHANICS    OF    ENGINEERING. 

to  increase  its  volume  sixfold  by  expanding  behind  a  piston. 
To  what  temperature  has  it  cooled  ?  Here  Tm  —  293°  Abs. 
Cent.,  the  ratio  Vm  I  Vn  =  £,  and  Tn  is  required.  Hence, 
from  (3), 

'     =  293  ^  v~*~=  119<5°  Abs- 


or  =  —  1540  Cent.,  indicating  extreme  cold. 

From  these  two  examples  the  principle  of  one  kind  of  ice- 
making  apparatus  is  very  evident.  As  to  the  work  necessary 
to  compress  the  air  in  Example  1,  see  §  483.  It  is  also  evident 
why  motors  using  compressed  air  expansively  have  to  encoun- 
ter the  difficulty  of  frozen  watery  vapor  (present  in  the  air  to 
some  extent). 

EXAMPLE  3.  —  What  is  the  tension  of  the  air  in  Example  1 
(suddenly  compressed  to  -J-  its  original  volume)  immediately 
after  the  compression,  if  the  original  tension  was  one  atmos- 
phere? That  is,  with  Vn  :  Vm  =  1  :  6,  and  j?w  =  14.7  Ibs.  per 
sq.  inch,  pn  =  2  From  eq.  (1),  (in.,  lb.,  sec.,) 


pn  =  14.7  X  6*  =  14.7  V2l6  =  216 

Ibs.  per  sq.  inch  ;  whereas,  if,  after  compression  and  without 
change  of  volume,  it  cools  again  to  20°  Cent.,  the  tension  is 
only  14.7  X  6  =  88.2  Ibs.  per  sq.  inch  (now  using  Mariotte's 
law). 

479.  Work  of  Expanding  Steam  following  Mariotte's  Law.  — 
Although  gases  do  not  in  general  follow  Mariotte's  law  in  ex- 
panding behind  a  piston  (without  special  provision  for  sup- 
plying heat),  it  is  found  that  the  tension  of  saturated  steam 
(i.e.,  saturated  at  the  beginning  of  the  expansion)  in  a  steam- 
engine  cylinder,  when  left  to  expand  after  the  piston  has 
passed  the  point  of  "  cut-af"  diminishes  very  nearly  in 
accordance  with  Mariotte's  law,  which  may  therefore  be  ap- 
plied in  this  case  to  find  the  work  done  per  stroke,  and  thence 
the  power.  In  Fig.  527  a  horizontal  steam-cylinder  is 


EXPANDING   STEAM. 


625 


— I- — 


shown  in  which  the  piston  is  making  its  left-to-right  stroke. 

The   "  back-pressure"   is  con- 

stant and  =  Fq,  F  being  the 

area  of  the  piston  and  q  the 

intensity  (i.e.,  per  unit  area) 

of  the  back  or  exhaust  pres- 

sure on  the  right  side  of  the 

piston  ;     while    the    forward 

pressure  on  the  left  face  of  the 

piston  =  Fp,  in  which  p  is  the 

steam-pressure  per  unit  area, 

and  is   different   at   different 

points  of  the  stroke.  While  the 

piston  is  passing  from  O"  to 

D"  ,p  is  constant,  being  —  pb  =  the  boiler-pressure,  since  the 

steam-port  is  still  open.     Between  D"  and  C"  ,  however,  the 

steam  being  cut  off  (i.e.,  the  steam-port  is  closed)  at  D"  ,  a  dis- 

tance a  from  0'  ',  p  decreases  with  Mariotte's  law  (nearly),  and 

its  value  is  (Fa  -r-  Fx)pb  at  any  point  on  C"Dfl  ',  x  being  the 

distance  of  the  point  from  0"  . 

Above  the  cylinder,  conceive  to  be  drawn  a  diagram  in 
which  an  axis  OX\§  \\  to  the  cylinder-axis,  OY  an  axis  1  to 
the  same,  while  0  is  vertically  above  the  left-hand  end  of  the 
cylinder.  As  the  piston  moves,  let  the  value  of  p  correspond- 
ing to  each  of  its  positions  be  laid  off,  to  scale,  in  the  vertical 
immediately  above  the  piston,  as  an  ordinate  from  the  axis  X. 
Make  OD'  =  q  by  the  same  scale,  and  draw  the  horizontal 
D'C'.  Then  the  effective  work  done  on  the  piston-rod  while 
it  moves  through  any  small  distance  dx  is 

d  W  =  force  X  distance  =  F(  p  —  q)dx, 

and  is  proportional  to  the  area  of  the  strip  RS,  whose  width  is 
dx  and  length  =  p  —  q  ;  so  that  the  effective  work  of  one 
stroke  is 


--q)dx,     .    .    .    (1) 


626  MECHANICS   OF  ENGINEERING. 

and  is  represented  graphically  by  the  area  A'ARBC'D'A'* 
From  0"  to  D"  p  is  constant  and  =  pb  (while  q  is  constant  at 
all  points),  and  x  varies  from  0  to  a  ; 

q)a,     .    .    (2) 

which  may  be  called  the  work  of  entrance,  and  is  represented 
by  the  area  of  the  rectangle  A'  ADD'. 

From  D"  to  Q  "  p  is  variable  and,  by  Mariotte's  law,  =  —  pb  ; 


.,-q(l-a)     ...    (3) 


=  the  work  of  expansion,  adding  which  to  that  of  entrance, 
we  have  for  the  total  effective  work  of  one  stroke 


By  effective  work  we  mean  that  done  upon  the  piston-rod 
and  thus  transmitted  to  outside  machinery.  Suppose  the 
engine  to  be  "  double-acting"  ;  then  at  the  end  of  the  stroke  a 
communication  is  made,  by  motion  of  the  proper  valves,  be- 
tween the  space  on  the  left  of  the  piston  and  the  condenser  of 
the  engine  ;  and  also  between  the  right  of  the  piston  and  the 
boiler  (that  to  the  condenser  now  being  closed).  On  the  return 
stroke,  therefore,  the  conditions  are  the  same  as  in  the  forward 
stroke,  except  that  the  two  sides  of  the  piston  have  changed 
places  as  regards  the  pressures  acting  on  them,  and  thus  the 
same  amount  of  effective  work  is  done  as  before. 

If  n  revolutions  of  the  fly-wheel  are  made  per  unit  of  time 
(two  strokes  to  each  revolution),  the  effective  work  done  per 
unit  of  time,  i.e.,  the  power  of  the  engine,  is 

Z  =  2n  W=  ZnF  [ap^l  +  log..  (1)1  -  ql^.     .    (5) 


WORK   OF   STEAM-ENGINES.  6 '27 

For  simplicity  the  above  theory  has  omitted  the  considera- 
tion of  " clearance"  that  is,  the  fact  that  at  the  point  of  " cut- 
off "  the  mass  of  steam  which  is  to  expand  occupies  not  only 
the  cylindrical  volume  Fa,  but  also  the  "  clearance"  or  small 
space  in  the  steam-passages  between  the  valve  and  the  entrance 
of  the  cylinder,  the  space  between  piston  and  valve  which  is 
never  encroached  upon  by  the  piston.  "  Wire-drawing"  has 
also  been  disregarded,  i.e.,  the  fact  that  during  communication 
with  the  boiler  the  steam-pressure  on  the  piston  is  a  little  less 
than  boiler-pressure.  For  these  the  student  should  consult 
special  works,  and  also  for  the  consideration  of  water  mixed 
with  the  steam,  etc.  Again,  a  strict  analysis  should  take  into 
account  the  difference  in  the  areas  which  receive  fluid-pressure 
on  the  two  sides  of  the  piston. 

EXAMPLE  1. — A  reciprocating  steam-engine  makes  120  revo- 
lutions per  minute,  the  boiler-pressure  is  40  Ibs.  by  the  gauge 
(i.e.,^>&  =  40  -f- 14.7=  54.7  Ibs.  per  sq.  inch),  the  piston  area 
is  F=  120  sq.  in.,  the  length  of  stroke  I  =  16  in.,  the  steam 
being  "cut  off"  at  J  stroke  (/.  a  =  4  in.,  and  I  :  a  =  4.00), 
and  the  exhaust  pressure  corresponds  to  a  "  vacuum  of  25 
inches"  (by  which  is  meant  that  the  pressure  of  the  exhaust 
steam  will  balance  5  inches  of  mercury),  whence  q  —  -f$  of 
14.7  =  2.45  Ibs.  per  sq.  inch.  Required  the  work  per  stroke, 
W,  and  the  corresponding  power  Z. 

Since  I  :  a  =  4,  we  have  log.e  4  =  2.302  X  .60206  =  1.386, 
and  from  eq.  (4),  (foot,  lb.,  sec.,) 

W=  |££  (54.7  X  144) .  j. .  [2.386]  -  -fjf  (2.45  X  144) .  f 

=  5165.86  -  392.0  =  4773.868  ft.  Ibs.  of    work  per  stroke, 
and  therefore  the  power  at  2  rev.  per  sec.  (eq.  5)  is 

Z  =  2  X  2  X  4773.87  =  19095.5  ft.  Ibs.  per  second. 
Hence  in  horse-powers,  which,  in  ft.,-lb.-sec.  system,  =  Z-i-550, 

Power  =  19095.5  -5-  550  =  34.7  H.  P. 
EXAMPLE  2. — Required  the  weight  of  steam  consumed  per 


628  MECHANICS   OF   ENGINEERING. 

second  by  the  above  engine  with  given  data ;  assuming  with 
Weisbach  that  the  heaviness  of  saturated  steam  at  a  definite 
pressure  (and  a  corresponding  temperature,  §  469)  is  about  f  of 
that  of  air  at  the  same  pressure  and  temperature. 

The  heaviness  of  air  at  54.7  Ibs.  per  sq.  in.  tension  and 
temperature  287°  Fahr.  (see  table,  §  469)  would  be,  from  eq. 
(12)  of  §  472  (see  also  §  409), 

_yj\     j>        0807X493    64.7 

T    '  po        460  +  287    '14.7 

Ibs.  per  cub.  foot,  f  of  which  is  0.1237  Ibs.  per  cub.  ft.  Now 
the  volume  of  steam,  of  this  heaviness,  admitted  from  the 
boiler  at  each  stroke  is  V  =  Fa  =  |f  £  .  £  =  0.2777  cub.  ft.? 
and  therefore  the  weight  of  steam  used  per  second  is 

4  X  .2777  X  0.1237  =  0.1374  Ibs. 

Hence,  per  hour,  0.1374  X  3600  =  494.6  Ibs.  of  feed-water 
are  needed  for  the  boiler. 

If,  with  this  same  engine,  the  steam  is  used  at  full  boiler 
pressure  throughout  the  whole  stroke,  the  power  will  be 
greater,  viz.  =  %nFl(pb  —  q)  —  33440  ft.  Ibs.  per  sec.,  but 
the  consumption  of  steam  will  be  four  times  as  great;  and 
hence  in  economy  of  operation  it  will  be  only  0.44  as  efficient 
(nearly). 

480.  Graphic  Representation  of  any  Change  of  State  of  a  Con- 
fined Mass  of  Gas. — The  curve  of  expansion  AB  in  Fig.  527  is 
an  equilateral  hyperbola,  the  axes  JTand  Y  being  its  asymp- 
totes. If  compressed  air  were  used  instead  of  steam  its  ex- 
pansion curve  would  also  be  an  equilateral  hyperbola  if  its 
temperature  could  be  kept  from  falling  during  the  expansion 
(by  injecting  hot-water  spray,  e.g.),  and  then,  following 
Mariotte's  law,  we  would  have,  as  for  steam,  (§  475,)  p  V=  con- 
stant, i.z.,pFx  =  constant,  and  therefore  px  =  constant,  which 
is  the  equation  of  a  hyperbola,  p  being  the  ordinate  and  x  the 
abscissa.  This  curve  (dealing  with  a  perfect  gas)  is  also  called 
an  isothermal,  the  x  and  y  co-ordinates  of  its  points  being  pro- 


GRAPHICS   OF   CHANGE   OF   STATE   OF   GAS. 


629 


X 


portional  to  the  volume  and  tension,  respectively,  of  a  mass  of 
air  (or  perfect  gas)  whose  temperature  is  maintained  constant. 
Hence,  in  general,  if  a  mass  of  gas  be  confined  in  a  rigid 
cylinder  of  cross-sec- 
tion F  (area),  provided 
with  an  air-tight  pis- 
ton, Fig.  528,  its  vol- 
ume, Fx,  is  propor- 
tional to  the  distance 
OD  =--  x  (of  the  piston 
from  the  closed  end  of 
the  cylinder)  taken  as 
an  abscissa,  while  its  o 
tension  p  at  the  same 
instant  may  be  laid  off 
as  an  ordinate  from  D. 
Thus  a  point  A  is  fixed.  Describe  an  equilateral  hyperbola 
through  A,  asymptotic  to  X  and  I7,  and  mark  it  with  the  ob- 
served temperature  (absolute)  of  the  air  at  this  instant.  In  a 
similar  way  the  diagram  can  be  filled  up  with  a  great  number 
of  equilateral  hyperbolas,  or  isothermal  curves,  each  for  its 
own  temperature.  Any  point  whatever  (i.e.,  above  the  critical 
temperature)  in  the  plane  angular  space  YOX  will  indicate  by 
its  co-ordinates  a  volume  and  a  tension,  while  the  correspond- 
ing absolute  temperature  T  will  be  shown  by  the  hyperbola 
passing  through  the  point,  since  these  three  variables  always 
satisfy  the  relation  (§  472) 


528. 


pV  .      pFx 

-Zj,-  =  const. ;  i.e.,  ^-  . 


.    .     .    (1)' 


Any  change  of  state  of  the  gas  in  the  cylinder  may  now  be 
represented  by  a  line  in  the  diagram  connecting  the  two  points 
corresponding  to  its  initial  and  final  states.  Thus,  a  point 
moving  along  the  line  AB,  a  portion  of  the  isothermal  marked 
293°  Abs.  Cent,  represents  a  motion  of  the  piston  from  D  to 
E,  and  a  consequent  increase  of  volume,  accompanied  by  just 
sufficient  absorption  of  heat  by  the  gas  (from  other  bodies)  to 
maintain  its  temperature  at  that  figure  (viz.,  its  temperature  at 


630  MECHANICS    OF    ENGINEERING. 

A).  If  the  piston  move  from  D  to  E,  without  transmission 
of  heat,  i.e.,  adiabatically  (§  478),  the  tension  falls  more 
rapidly,  and  a  point  moving  along  the  line  AB1  represents  the 
corresponding  continuous  change  of  state.  AB'  is  a  portion 
of  an  adiabatic  curve,  whose  equation,  from  §  478,  is 

iL  =  [llfj   '     °r    1**  =#***?  =  <**&.;     •    00 

in  which  pK  and  XK  refer  to  the  point  K  where  this  particular 
adiabatic  curve  cuts  the  isothermal  of  freezing-point.  Evi- 
dently an  adiabatic  may  be  passed  through  any  point  of  the 
diagram.  The  mass  of  gas  in  the  cylinder  may  change  its 
state  from  A  to  B'  by  an  infinite  number  of  routes,  or  lines  on 
the  diagram,  the  adiabatic  route,  however,  being  that  most  likely 
to  occur  for  a  rapid  motion  of  the  piston.  For  example,  we 
may  cool  it  without  allowing  the  piston  to  move  (and  hence 
without  altering  its  volume  Xor  the  abscissa  x)  until  the  pres- 
sure falls  to  a  value  pB>  =  DL  =  EB\  and  this  change  is  rep- 
resented by  the  vertical  path  from  A  to  L ;  and  then  allow  it 
to  expand,  and  push  the  piston  from  D  to  E  (i.e.,  do  external 
work),  during  which  expansion  heat  is  to  be  supplied  at  just 
such  a  rate  as  to  keep  the  tension  constant,  ^=pp>  =PLI  this 
latter  change  corresponding  to  the  horizontal  path  LB'  from 
L  to  Bf. 

It  is  further  noticeable  that  the  work  done  by  the  expanding 
gas  upon  the  near  face  of  the  piston  (or  done  upon  the  gas  when 
compressed)  when  the  space  dx  is  described  by  the  piston,  is 
=  Fpdx,  and  therefore  is  proportional  to  the  area  pdx  of  the 
small  vertical  strip  lying  between  the  axis  X  and  the  line  or 
route  showing  the  change  of  state ;  whence  the  total  work  done 
on  the  near  piston-face,  being  =  Ffpdx,  is  represented  by  the 
area  fpdx  of  the  plane  figure  between  the  initial  and  final 
ordinates,  the  axis  X  and  the  particular  route  followed  be- 
tween the  initial  and  final  states  (K.B.  We  take  no  account 
here  of  the  pressure  on  the  other  side  of  the  piston,  the  latter 
depending  on  the  style  of  engine).  For  example,  the  work 
done  on  the  near  face  of  the  piston  during  adiabatic  expansion 


WORK   OF  ADIABATIC   EXPANSION. 


631 


from  D  to  E  is  represented  by  the  plane  figure  AB'EDA, 
and  is  measured  by  its  area. 

The  mathematical  relations  between  the  quantities  of  heat 
imparted  or  rejected  by  conduction  and  radiation,  and  trans- 
formed into  work,  in  the  various  changes  of  which  the  con- 
fined gas  is  capable,  belong  to  the  subject  of  Thermodynamics, 
which  cannot  be  entered  upon  here. 

It  is  now  evident  how  the  cycle  of  changes  which  a  definite 
mass  of  air  or  gas  experiences  when  used  in  a  hot-air  engine, 
compressed-air  engine,  or  air-compressor,  is  rendered  more  in- 
telligible by  the  aid  of  such  a  diagram  as  Fig.  528 ;  but  it 
must  be  remembered  that  during  the  entrance  into,  or  exit 
from,  the  cylinder,  of  the  mass  of  gas  used  in  one  stroke,  the 
distance  x  does  not  represent  its  volume,  and  hence  the  locus 
of  the  points  in  the  diagram  determined  by  the  co-ordinates  p 
and  x  during  entrance  and  exit  does  not  indicate  changes  of 
state  in  the  way  just  explained  for  the  mass  when  confined  in 
the  cylinder.  However,  the  work  done  by  or  upon  the  gas 
during  entrance  and  exit  will  still  be  represented  by  the  plane 
figure  included  by  that  locus  (usually  a  straight  horizontal 
line,  pressure  constant)  and  the  axis  of  X  and  the  terminal 
ordinates. 

481.  Adiabatic  Expansion  in  an  Engine  using  Compressed  Air. 
— Fig.  529.  Let  the  compressed  air  at  a  tension  pm  and  an 
absolute  temperature  Tm  be  supplied 
from  a  reservoir  (in  which  the  loss  is 
continually  made  good  by  an  air-corn- 
pressor).  Neglecting  the  resistance  of 
the  port,  its  tension  and  temperature 
when  behind  the  piston  are  still  pm  and 
Tm .  Let  xn  =  length  of  stroke,  and  o 
let  the  cut-off  (or  closing  of  communi- 
cation with  the  reservoir)  be  made  at 
some  point  D  where  x  —  xm ,  the  posi- 
tion of  D  being  so  chosen  (i.e.,  the 
ratios?  :  xn  so  computed)  that  after  adia- 

,     , .  .          -  T\  j.       TV  A*  FIG.  529. 

batic  expansion  from  D  to  E  the  pres- 
sure shall  have  fallen  fromjpm  at  M  (state  m)  to  a  value  pn  = 


632  MECHANICS   OF  ENGINEERING. 

=  one  atmosphere  at  JV(state  n\  at  the  end  of  stroke  ;  so  that 
when  the  piston  returns  the  air  will  be  expelled  ("exhausted") 
at  a  tension  equal  to  that  of  the  external  atmosphere  (though 
at  a  low  temperature).  Hence  the  back-pressure  at  all  points 
either  way  will  be  =  pn  per  unit  area  of  piston,  and  hence  the 
total  back-pressure  =  Fpn,  F  being  the  piston  area. 

From  0  to  D  the  forward  pressure  is  constant  and  =  Fpn  . 
and  the  effective  work,  therefore,  or  work  on  piston-rod  from 
O  to  D,  is 

W  =  F[pm  —  Pn\®m)          •       *       (1). 

represented  by  the  rectangle  M'MLN'.  The  cut-off  being 
made  at  D,  the  volume  of  gas  now  in  the  cylinder,  viz., 
Ym  =  Fxm,  is  left  to  expand.  Assuming  no  device  adopted 
(such  as  injecting  hot-water  spray)  for  preventing  the  cooling 
and  rapid  decrease  of  tension  during  expansion,  the  latter  is- 
adiabatic,  and  hence  the  tension  at  any  point  P  between  M 
and  N  will  be 


P=pm  *    .    [see  §478;   V=JM];  .    .  (ay 

.:   Work  of  expansion 

-pn}dx  =  Ff*pdx  -  Fpn(xn  -  O,  (2) 

m 

I 

and  is  represented  by  the  area  MPNL. 

i        /I  \t 


Now  substitute  (3)  in  (2)  and  then  add  (2)  to  (1),  noting  that 


COMPEESSED-AIR  ENGINE.  633 


which  furthermore,  since  n  and  m  are  adiabatically  related 
[see  (0)],  can  be  reduced  to 


and  we  have  finally  : 

Total  work  on  piston-  \       ^      Q  ^          I",       /ajw\*~| 
rod  per  stroke          \  ~~          =  8J^«  [         ^  J'      * 

But  Fxm  =  Vm,  and  the  adiabatic  relation  holds  good, 


e 


therefore  we  may  also  write 

TT=  3  FMJt>l  -?;    ....    (5) 


in  which  Vm  =  the  volume  which  the  mass  of  air  used  per 
stroke  occupies  in  the  state  m,  i.e.,  in  the  reservoir,  where  the 
tension  is^?m  and  the  absolute  temperature  =  Tm. 

To  find  the  work  done  per  pound  of  air  used  (or  other  unit 
of  weight),  we  must  divide  W  by  the  weight  G  =  Vmym  of 
the  air  used  per  stroke,  remembering  (eq.  (13),  §  472)  that 

Vmym  =  [  V»pmy.T.]  *  (Tmp.). 

Work  per  unit  of  weight  of\_»T  p,  H  _(Pn^r\  /g\ 
air  used  in  adiabatic  working  \  '  m  yoT\_  \pm)  J 


The  back-pressure pn  =pa  =  one  atmosphere. 
In  (6)  y0  =  .0807  Ibs.  per  cub.  foot,  p0  —  14.7  Ibs.  per  sq 
inch,  and  T0  =  273°  Abs.  Cent,  or  492°  Abs.  Fahr. 


634  MECHANICS   OF  ENGINEERING. 

It  is  noticeable  in  (6)  that  for  given  tensions  pm  and  pn)  the 
work  per  unit  of  weight  of  air  used  is  proportional  to  the  ab- 
solute temperature  Tm  of  the  reservoir.  The  temperature  Tm 
to  which  the  air  has  cooled  at  the  end  of  the  stroke  is  obtained 
as  in  Example  2,  §  478,  and  may  be  far  below  freezing-point 
unless  Tm  is  very  high  or  the  ratio  of  expansion,  xm  :  xn  ,  large. 

EXAMPLE.  —  Let  the  cylinder  of  a  compressed-air  engine  have 
a  section  of  F  =  108  sq.  in.  and  a  stroke  xn  =  15  inches.  The 
compressed  air  entering  the  cylinder  is  at  a  tension  of  2  atmos. 
(i.e.,  pm  =  29.4  Ibs.  per  sq.  in.,  and  pn  -±-pm  =  -J),  and  at  a 
temperature  of  27°  Cent,  (i.e.,  Tm  =  300°  Abs.  Cent.).  Ke- 
quired  the  proper  point  of  cut-off,  or  a?m  =  ?  ,  in  order  that  the 
tension  may  fall  to  one  atmosphere  at  the  end  of  the  stroke  ; 
also  the  work  per  stroke,  and  the  work  per  pound  of  air.  Use 
the/002,  pound,  and  second. 

From  eq.  (a),  above,  we  have 

®m=xn  f^§=  1*354  /  T  =  0.7875  ft.  =  9.45  inches, 
W  V    4 

and  hence  the  volume  of  air  in  state  m,  used  per  stroke  [eq. 

(5)]  is 

Vm  =  Fxm  =  |f|  X  0.7875  =  0.5906  cubic  feet  ; 
while  the  work  per  stroke  is 

W  =  3  X  0.5906  X  29.4  X  144  X  [1  -  (i)*  ]  =  1545  ft.  Ibs., 
and  the  work  obtained  from  each  pound  of  air,  eq.  (6), 


ft.  Ibs.  per  pound  of  air  used. 

The  temperature  to  which  the  air  has  cooled  at  the  end  of 
stroke  [eq.  (2),  §  478]  is 


COMPKESSED-AIR   ENGINE.  635 


Tn  =  Tm  :    &-  =  300  X  V  i  =  300  X  .794  =  238°  Abs.  C. ; 

i.e.,  —  35°  Centigrade. 

482.  Remarks  on  the  Preceding. — This  low  temperature  is 
objectionable,  causing,  as  it  does,  the  formation  and  gradual 
accumulation  of  snow,  from  the  watery  vapor  usually  found 
in  small  quantities  in  the  air,  and  the  ultimate  blocking  of  the 
ports.  By  giving  a  high  value  to  Tm ,  however,  i.e.,  by  heat- 
ing the  reservoir,  Tn  will  be  correspondingly  higher,  and  also 
the  work  per  pound  of  air,  eq.  (6).  If  the  cylinder  be  encased 
in  a  "  jacket"  of  hot  water,  or  if  spray  of  hot  water  be  injected 
behind  the  piston  during  expansion,  the  temperature  may  be 
maintained  nearly  constant,  in  which  event  Mariotte's  law  will 
hold  for  the  expansion,  and  more  work  will  be  obtained  per 
pound  of  air;  but  the  point  of  cut-off  must  be  differently 
placed.  Thus  if,  in  eq.  (4),  §  479,  we  make  the  back-pressure, 
which  =  (Fa  -s-  Fl)pb ,  equal  to  the  value  to  which  the  air- 
pressure  has  fallen  at  the  end  of  the  stroke  by  Mariotte's  law, 
we  have 

Work  per  stroke  with  1  __  c*       i        (l\_  Y      1        (^\   C[\ 
isotherm,  expans.    f  ~~     a^b  °&'  \a/~      b^b  °£'«  \^"/>  v 

and  hence 

Work  per  unit  of  weight  of  air,  \  _  j,     p0     i        /_£  \       ,~ 
with  isothermal  expansion         )          m  y0T0     ^'*  v»/ 

Applying  these  equations  to  the  data  of  the  example,  we 
obtain 

Work  per  unit  of  weight  of  air  with  iso-  \  _  Q  go  y1    P°    . 
thermal  expansion  )  "  my,T\ ' 

whereas,  with  adidbatic  expansion,  work  I  _  Q  gg  T7    P» 
per  unit  of  weight  of  air  is  only  }  "  "V0^V 


636  MECHANICS   OF   ENGINEERING. 

483.  Double-acting  Air-compressor,  with  Adiabatic  Compres- 
sion.— This  is  the  converse  of  §  481.  In  Fig.  530  we  have  the 
piston  moving  from  right  to  left,  compressing  a  mass  of  air 
which  at  the  beginning  of  the  stroke  fills  the  cylinder.  This  is 
brought  about  by  means  of  an  external 
motor  (steam-engine  or  turbine,  e.g.) 
which  exerts  a  thrust  or  pull  along  the 
piston-rod,  enabling  it  with  the  help 
of  the  atmospheric  pressure  of  the 
fresh  supply  of  air  flowing  in  behind 
•x  it,  to  first  compress  a  cylinder-full  of 
air  to  the  tension  of  the  compressed 
p  air  in  the  reservoir,  and  then,  the 
Lj  port  or  valve  opening  at  this  stage, 
to  force  or  deliver  it  into  the  reservoir. 
Let  the  temperature  and  tension  of  the 


cylinder-full  of  fresh  air  be  TUl  and 

pni ,  and  the  tension  in  the  reservoir  be  pmi .  Suppose  the 
compression  adiabatic.  As  the  piston  passes  from  E  toward 
the  left,  the  air  on  the  left  has  no  escape  and  is  compressed,  its 
tension  and  temperature  increasing  adiabatically  until  it  reaches 
a  value  pmi  =  that  in  reservoir,  at  which  instant,  the  piston 
being  at  some  point  D,  a  valve  opens  and  the  further  progress 
of  the  piston  simply  transfers  the  compressed  air  into  the  re- 
servoir without  further  increasing  its  tension.  Throughout 
the  whole  stroke  the  piston-rod  has  the  help  of  one  atmosphere 
pressure  on  the  right  face,  since  a  new  supply  of  air  is  entering 
on  the  right  to  be  compressed  in  its  turn  on  the  return  stroke. 
The  work  done  from  E  to  D  may  be  called  the  work  of  com- 
pression •  that  from  D  to  0,  the  work  of  delivery. 

[Since,  here,  dx  and  dW(or  increment  of  work)  have  con- 
trary signs,  we  introduce  the  negative  sign  as  shown.] 

rD 

The  work  of  compression  =  — J EF(p  —  pn^)dx.  .     .     .     (Ic) 
The  work  of  delivery       = —Pm.—pn^dx.     .    .     (Id) 


AIR-COMPKESSOK.  637 

In  these  equations  only^>  and  x  are  variables.  In  the  sum- 
mation indicated  in  (1<?)  p  changes  adiabatically ;  in  (l.d)p  is 
constant  =pmi  as  now  written. 

In  the  adiabatic  compression  the  air  passes  from  the  state  n^ 
to  the  state  m^  (see  JV,  and  Ml  in  figure). 

The  summations  in  these  equations  being  of  the  same  form 
as  those  in  equations  (1)  and  (2)  of  §  481,  but  with  limits  in- 
verted, we  may  write  immediately, 

Work  per  stroke  =  W=  3  F«J>TOI  |~1  —  (— Y~|  '  •    (2) 

\prnj  J 

and 

Work  per  unit  of  weight )  _.  077      PO    F-,  _(PnL\*~]         /o\ 
o/"  o*>  compressed          j  "         mi  ^  j7  [_        \pm  I 

The  value  of  T7^ ,  at  the  immediate  end  of  the  sudden  com- 
pression, by  eq.  (2)  of  §  (478),  is 


(4) 


The  temperature  of  the  reservoir  being  Tm  ,  as  in  §  481 
(usually  much  less  than  Tmi  ),  the  compressed  air  entering  it 
cools  down  gradually  to  that  temperature,  Tm  ,  contracting  in 
volume  correspondingly  since  it  remains  at  the  same  tension 
pmi  .  The  mechanical  equivalent  of  this  heat  is  lost. 

Let  us  now  inquire  what  is  the  efficiency  of  the  combination 
of  air-compressor  and  compressed-air  engine,  the  former  sup- 
plying air  for  the  latter,  both  working  adiabatically,  assuming 
that  no  tension  is  lost  by  the  compressed  air  in  passing  along 
the  reservoir  between,  i.e.,  that  pmi  =  pm  .  Also  assume  (as 
already  implied,  in  fact)  that  j?Wl  =pn  =  one  atmos.,  and  that 
the  temperature,  Tn^  of  the  air  entering  the  compressor  cyl- 
inder is  equal  to  that,  Tm  ,  of  the  reservoir  and  transmission- 
pipe. 

To  do  this  we  need  only  find  the  ratio  of  the  amount  of 
work  obtained  from  one  pound  (or  other  unit  of  weight)  in  the 
compressed-air  engine  to  the  amount  spent  in  compressing  one 
pound  of  air  in  the  compressor.  Calling  this  ratio  t?,  the 


638     -  MECHANICS   OF  ENGINEERING. 

efficiency,  and  dividing  eq.  (6)  of  §  481  by  eq.  (3)  of  this  para- 
graph, we  have,  with  substitutions  just  mentioned, 

_  Tm     _  Abs.  temp,  of  outer  free  air  0  ,_ 


_  _ 

of  sudden  compression, 


Tmi         (  Abs.  temp,  of  air  at  end 
(      of  sudd 


or,  substituting  from  eq.  (4),  and  remembering  that  Tni  =  T 
we  have  also 


(6) 

also,  since 


T 

^  ...  -*•  m 

we  may  write 

_  Tn  __  Ab.  tern,  air  leaving  eng.  cyl.  ,^ 

Tm  Ab.  tern,  outer  free  air. 

For  practical  details  of  the  construction  and  working  of 
engines  and  compressors,  and  the  actual  efficiency  realized,  the 
student  may  consult  special  works,  as  they  lie  somewhat  be- 
yond the  scope  of  the  present  work. 

EXAMPLE  1. — In  the  example  of  §  445,  the  ratio  of  pm  to pn 
was  —  £.  Hence,  if  compressed  air  is  supplied  to  the  reser- 
voir under  above  conditions,  the  efficiency  of  the  system  is, 
from  eq.  (6),  77  =  V^  —  0.794,  about  80  per  cent. 

7}         1 

EXAMPLE  2. — If  the  ratio  of  the  tensions  is  as  small  as  —  =  — , 

Pm         6 

the  efficiency  would  be  only  (-J-)*  =  0.55 ;  i.e.,  45  per  cent  of 
the  energy  spent  in  the  compressor  is  lost  in  heat. 

EXAMPLE  3. — What  horse-power  is  required  in  a  blowing 
engine  to  furnish  10  Ibs.  of  air  per  minute  at  a  pressure  of 
4  atmos.,  with  adiabatic  compression,  the  air  being  received 
by  the  compressor  at  one  atmosphere  tension  and  27°  Cent, 
(ft.-lb.-sec.  system).  Since  27°  C.  =  300°  Abs.  C.  =  Tni ,  we 
have,  from  eq.  (4), 

Tmi  =  300  (f )»  =  477°  Abs.  Cent. ; 
and  hence,  eq.  (3), 


HOT-AIB  ENGINES.  630 

1 4  7  v  1 44  F          /I  Yin 
TteworJcper  pound  of  air  =  3X  477^^  ^[l  -  g) 

=  50870  ft.  Ibs.  per  pound  of  air.  Hence  10  Ibs.  of  air  will 
require  508700  ft.  Ibs.  of  work ;  and  if  this  is  done  every  min- 
ute we  have  the  req.  H.  P.  =  *££f£f-  =  15A  H'  R 

NOTE. — If  the  compression  could  be  made  isothermal,  an 
approximation  to  which  is  obtained  by  injecting  a  spray  of 
cold  water,  we  would  have,  from  eqs.  (1)  and  (2)  of  §  482 : 

Workper  \_T     p0  ,       / #»,  \     300  X  14.7  X  144 

II.  air     \  -T^f0 10g'efc)  =       .0807  X  273      X  l^ 

=  39950  ft.  Ibs.  per  lb.,  and  the  corresponding  H.  P.  =  12.1 ; 
a  saving  of  about  25  per  cent,  compared  with  the  former. 
The  difference  was  employed  in  heating  the  air  in  the  air-com- 
pressor with  adiabatic  compression,  and  was  lost  when  that 
extra  heat  was  dissipated  in  the  reservoir  as  the  air  cooled 
again.  This  difference  is  easily  shown  graphically  by  compar- 
ing in  the  same  diagram  the  areas  representing  the  work  done 
in  the  two  cases. 

484.  Hot-air  Engines. — Since  we  have  seen  that  the  tension 
of  air  and  other  gases  can  be  increased  by  heating,  if  the  vol- 
ume be  kept  the  same,  a  mass  of  air  thus  treated  can  after- 
wards be  allowed  to  expand  in  a  working  cylinder,  and  thus 
become  a  means  of  converting  heat  into  work.  In  Stirling^ 
hot-air  engine  a  definite  confined  mass  of  air  is  used  indefinitely 
without  loss  (except  that  occasional  small  supplies  are  needed 
to  make  up  for  leakage),  and  is  alternately  heated  and  cooled. 
A  displacement-plunger,  or  piston,  fitting  loosely  in  a  bell-like 
chamber,  is  so  connected  with  the  piston  of  the  working 
cylinder  and  the  fly-wheel,  that  its  forward  stroke  is  made 
while  the  other  piston  waits  at  the  beginning  of  its  stroke. 
In  this  motion  the  plunger  causes  the  confined  air  to  pass  in  a 
thin  sheet  over  the  top  and  sides  of  the  furnace  dome,  thus 
greatly  increasing  its  tension.  The  air  then  expands  behind 
the  working  piston  with  falling  tension  and  temperature,  and, 


640  MECHANICS   OF   ENGINEERING. 

while  that  piston  pauses  at  the  end  of  its  forward  stroke,  is 
again  shifted  in  position,  though  without  change  of  volume, 
by  the  return  stroke  of  the  plunger,  in  such  a  way  as  to  pass 
through  a  coil  of  pipes  in  which  cold  water  is  flowing.  This 
reduces  both  its  temperature  and  tension,  and  hence  its  resist- 
ance to  the  piston  on  the  return  stroke  is  at  first  less  than  at- 
mospheric, but  is  gradually  increased  by  the  compression. 
This  cycle  of  changes  is  repeated  indefinitely,  and  is  easily 
traced  on  a  diagram  like  that  in  Fig.  528,  and  computations 
made  accordingly. 

A  special  invention  of  Stirling's  is  the  "  regenerator"  or  box 
filled  with  numerous  sheets  of  wire  gauze,  in  its  passage 
through  which  the  working  air,  after  expansion,  deposits  some 
of  its  heat,  which  it  re-absorbs  to  some  extent  when,  after 
further  cooling  in  the  "  refrigerator"  or  pipe  coil  and  com- 
pression by  the  return  stroke  of  the  piston,  it  is  made  to  pass 
backward  through  the  regenerator  to  be  further  heated  by  the 
furnace  in  readiness  for  a  forward  stroke.  This  feature,  how- 
ever, has  not  realized  all  the  expectations  of  its  inventor  and 
improvers,  as  to  economy  of  heat  and  fuel. 

In  Ericsson? s  hot-air  engine,  of  more  recent  date,  the  dis- 
placement-plunger fits  its  cylinder  air-tight,  but  valves  can  be 
opened  through  its  edges  when  moving  in  one  direction,  thus 
causing  it  to  act  temporarily  as  a  loose  plunger,  or  shifter. 
The  two  pistons  move  simultaneously  in  the  same  direction  in 
the  same  cylinder,  but  through  different  lengths  of  stroke,  so 
that  the  space  between  them  is  alternately  enlarged  and  con- 
tracted. The  working  piston  also  has  valves  opening  through 
it  for  receiving  a  fresh  supply  of  air  into  the  space  between 
the  two  pistons.  During  the  forward  stroke  a  fresh  instal- 
ment from  the  outer  air  enters  through  the  working  piston  into 
the  space  between  it  and  the  other,  whose  valves  are  now 
closed  and  which  is  now  expelling  from  its  further  face, 
through  proper  valves,  the  air  used  in  the  preceding  stroke ; 
no  work  is  done  in  this  stroke.  On  the  return  stroke  this 
fresh  supply  of  air  is  free  to  expand  behind  the  now  retreating 
working  piston,  while  its  tension  is  greatly  increased  by  its 
being  shifted  (at  least  a  large  portion  of  it)  over  the  furnace 


GAS-ENGINES.  641 

dome  through  the  valves  (now  open)  of  the  plunger  piston,  by 
the  motion  of  the  latter,  which  now  acts  as  a  loose  plunger. 
The  engine  is  therefore  only  single-acting,  no  work  being  done 
in  each  forward  stroke.  For  further  details,  see  Goodeve's 
and  Rankine's  works  on  the  steam-engine;  also  the  article 
"  Hot-air  Engine"  in  Johnson's  Cyclopedia  by  Fres.  Barnard, 
and  Rontgen's  Thermodynamics. 

485.  Brayton's  Petroleum-engine. — Although  a  more  recent 
invention  than  the  gas-engines  to  be  mentioned  presently,  this 
motor  is  more  closely  related  to  hot-air  engines  than  the  latter. 
By  a  slow  combustion  of  petroleum  vapor  the  gaseous  products 
of  combustion,  while  under  considerable  tension,  are  enabled 
to  follow  up  a  piston  with  a  sustained  pressure,  being  left  to 
expand  through  the  latter  part  of  the  stroke.     Thus  we  have 
the  furnace  and  working  cylinder  combined  in  one.     The 
gradual  combustion  is  accomplished  by  making  use  of  the 
principle  of  the  Davy  safety-lamp  that  flame  will  not  spread 
through  layers  of  wire  gauze  of  proper  fineness. 

486.  Gas-engines. — We  again  have  the  furnace  and  working 
cylinder  in  one  in  a  " gas-engine"  where  illuminating  gas  and 
atmospheric  air  are  introduced  into  the  working  cylinder  in 
proper  proportions  (about  ten  parts  of  air  to  one  of  gas,  by 
weight)  to  form  an  explosive  mixture  of  more  or  less  violence  and 
exploded  at  a  certain  point  of  the  stroke,  causing  a  very  sudden 
rise  of  temperature  and  tension',  after  which  the  mass  expands 
behind  the  piston  with  falling  pressure.     On  the  return  stroke 
the  products  of  combustion  are  expelled,  and  no  work  done, 
these  engines  being  single-acting.     In  some  forms  the  mixture 
is  compressed  before  explosion,  since  it  has  been  found  that 
under  this  treatment  a  mixture  containing  a  larger  proportion 
of  air  to  gas  can  be  made  to  ignite,  and  that  then  the  resulting 
pressure  is  more  gradual  and  sustained,  like  that  of  steam  or  of 
the  mixture  in  the  Brayton  engine.     That  is,  the   effect  is 
analogous  to  that  of  "  slow-burning  powder"  in  a  gun. 

In  the  "  Otto  Silent  Gas-engine"  the  explosion  occurs  only 
every  fourth  stroke,  and  one  side  of  the  piston  is  always  open 


642  MECHANICS    OF   ENGINEERING. 

to  the  air.  The  action  on  the  other  side  of  the  piston  is  as 
follows :  (1)  In  the  forward  stroke  a  fresh  supply  of  explosive 
mixture  is  drawn  into  the  cylinder  at  one  atmosphere  tension. 
(2)  The  next  (backward)  stroke  compresses  the  mixture  into 
about  one  fourth  of  its  original  bulk,  this  operation  occurring 
at  the  expense  of  the  kinetic  energy  of  the  fly-wheel.  (3)  The 
mixture  is  ignited,  the  pressure  rises  to  6  or  7  atmospheres, 
and  work  is  done  on  the  piston  through  the  next  (forward) 
stroke,  the  tension  of  the  products  of  combustion  having 
fallen  to  one  atmosphere  (nearly)  at  the  end  of  the  stroke. 
(4)  In  the  next  (backward)  stroke  the  products  of  combus- 
tion are  expelled  arid  no  work  is  done. 

The  Atkinson  "  Cycle  Gas-engine,"  an  English  invention  of 
recent  .date  (see  the  London  Engineer  for  May  1887 ;  pp.  361 
and  380)  also  makes  an  explosion  every  fourth  stroke,  but  the 
link  work  connecting  the  piston  and  fly-wheel  is  of  such  de- 
sign that  the  latter  makes  but  one  revolution  during  the  four 
strokes.  Also  the  length  of  the  expansion  or  working  stroke 
is  greater  than  that  of  the  compression  stroke  and  the  products 
of  combustion  are  completely  expelled.  Consequently  the  effi- 
ciency of  this  motor  is  at  present  greater  than  that  of  any  other 
gas-engine.  See  §  487. 

One  of  the  most  simple  gas-engines  is  made  by  the  Economic 
Motor  Company  of  New  York.  The  piston  has  no  packing, 
being  a  long  plunger  ground  to  fit  the  cylinder  accurately  and 
kept  well  lubricated.  As  with  most  gas-engines  the  cylinder  is 
encased  in  a  water-jacket  to  prevent  excessive  heating  of  the 
working  parts  and  consequent  decomposition  of  the  lubricant. 

For  further  details  on  these  motors,  see  Kankine's  "  Steam- 
engine,"  Clark's  "Gas-engines"  in  Yan  Nostrand's  Science 
Series,  and  article  "  Gas-engine"  in  Johnson's  Cyclopaedia ;  also 
Prof.  Thurston's  report  on  Mechanical  Engineering  at  the 
Vienna  Exhibition  of  1873,  and  proceedings  of  the  "  Society 
of  Engineers"  (England)  for  1881. 

487.  Efficiency  of  Heat-engines. — According  to  the  mechan- 
ical theory  of  heat,  the  combustion  of  one  pound  of  coal,  pro- 


GAS-ENGINES.  643 

ducing,  as  it  does,  about  14,000  heat-units  (British  Thermal 
units ;  see  §  149,  Mechanics)  should  furnish 

14,000  X  772  =  10,808,000  ft.  Ibs.  of  work, 

if  entirely  converted  into  work.  Let  us  see  how  nearly  this  is 
accomplished  in  the  performance  of  the  most  recent  and 
economical  marine  engines  of  the  present  day,  viz.,  the  triple 
expansion  engines  of  some  Atlantic  steamers,  which  are  claimed 
to  have  consumed  per  hour  only  1.25  Ibs.  of  coal  for  each 
measured  (u  indicated  ")  horse-power  of  effective  work  done  in 
their  cylinders.  The  work-equivalent  of  1.25  Ibs.  of  coal  per 
hour  is 

1.25  X  14,000  X  772  =  13,510,000  ft.  Ibs.  per  hour; 

while  the  actual  work  per  hour  implied  in  "  one  H.  P.  per 
hour"  is 

33000  X  60  =  1,980,000  ft.  Ibs.  per  hour. 

That  is,  the  engines  utilize  only  one  seventh  of  the  heat  of  com- 
bustion of  the  fuel. 

According  to  Prof.  Thurston,  this  is  a  rather  extravagant 
claim  (1.25),  the  actual  consumption  having  probably  been  1.4 
Ibs.  of  coal  per  H.  P.  per  hour. 

The  ordinary  compound  marine  engine  is  stated  to  use  as 
little  as  2.00  Ibs.  per  hour  for  each  H.  P. 

Most  of  the  heat  not  utilized  is  dissipated  in  the  condenser. 

Similarly,  the  water-jacket,  a  necessary  evil  in  the  operation 
of  the  gas-engine,  is  a  source  of  great  loss  of  heat  and  work. 
Still,  Mr.  Wm.  Anderson  in  his  recent  work,  "  Conversion  of 
Heat  into  Work"  (London,  1887),  mentions  a  motor  of  this 
class  as  having  converted  into  work  -J-  of  the  heat  of  combus- 
tion [an  Otto  "Silent  Gas-engine,"  tested  at  the  Stevens 
Institute,  Hoboken,  1ST.  J.,in  1883]  ;  while  Prof.  Unwin  found 
the  Atkinson  engine  (see  last  paragraph)  capable  of  returning 
(in  the  cylinder)  fully  -^  of  the  heat-equivalent  of  the  gas  con- 
sumed. [This  latter  result  was  confirmed  in  Philadelphia  in 
Jan.  1889  by  Prof.  Barr,  under  direction  of  Prof.  Thurston.] 


644  MECHANICS    OF   ENGINEERING. 

488.  Duty  of  Pumping-engines.  —  Another  way  (often   used 
in  speaking  of  the  performance  of  pumping-enginesj  of  ex- 
pressing the  degree  of  economy  attained  in  the  use  of  fuel  by 
the  combined  furnace,  boiler,  and  engine  is  to  give  the  num- 
ber of  foot-pounds  of  work  obtained  from  each  100  Ibs.  of  coal 
consumed  in  the  furnace,  calling  it  the  "  duty"  of  the  engine. 

For  example,  by  a  duty  of  99,000,000  ft.  Ibs.  it  is  meant  that 
from  each  pound  of  coal  990,000  ft.  Ibs.  of  work  are  obtained. 
From  this  we  gather  that,  since  one  horse-power  consists  of 
33,000  X  60  =  1,980,000  ft.  Ibs.  per  hour,  the  engine  men- 
tioned must  use  each  hour 

1,980,000  -f-  990,000  =  2  Ibs.  of  coal  for  each  H.  P.  developed  ; 

which  is  as  low  a  figure  as  that  attained  by  the  marine  engines 
last  quoted. 

489.  Buoyant  Effort  of  the  Atmosphere,  —  In  the  case  of  a 
body  of  large  bulk  but  of  small  specific  gravity  the  buoyant 
effort  of  the  air  (due  to  the  same  cause  as  that  of  water,  see 
§  456)   becomes   quite   appreciable,   and  may   sometimes  be 
greater  than  the  weight  of  the  body.     This  buoyant  effort  is 
equal  to  the  weight  of  air  displaced,  i.e.,  =  Yy,  where  Y  is 
the  volume  of  air  displaced,  and  y  its  heaviness. 

If  Gl  —  total  weight  of  the  body  producing  the  displace- 
ment, the  resultant  vertical  force  is 


(i) 


and  for  equilibrium,  or  suspension  in  the  air,  we  must  Lave 
P  =  0,  i.e., 

G,=  Yy  ........    (2) 

We  may  therefore  find  approximately  the  elevation  where 
a  given  balloon  will  cease  to  ascend,  by  determining  the  heavi- 
ness y  of  the  air  at  that  elevation  from  eq.  (2)  ;  then,  know- 
ing approximately  the  temperature  of  the  air  at  that  elevation, 
we  may  compute  its  tension  p  [eq.  (13),  §  472],  and  finally, 
from  eqs.  (3),  (4),  or  (5)  of  §  477,  obtain  the  altitude  required. 

EXAMPLE.  —  The  car  and  other  solid  parts  of  a  balloon  weigh 


BALLOONS.  645 

400  Ibs.,  and  the  bag  contains  12,000  cub.  feet  of  illuminating 
gas  weighing  0.030  Ib.  per  cub.  foot  at  a  tension  of  one  at- 
mosphere and  temperature  of  15°  Cent.,  so  that  its  total 
weight  =12,000  X  0.030  =  360  Ibs. 

Hence  G-l  =  760  Ibs.     We  may  also  write  with  sufficient 
accuracy  :  Whole  volume  of  displacement  =V=  12,000  cub.  ft. 

As  the  balloon  ascends  the  exterior  pressure  diminishes,  and 

the  confined  gas  tends  to  expand  and  so  in-  -M-O_ '     ^ 

crease  the  volume  of  displacement  V\  but  •  •-••--•    ••'•' 

this  we  shall    suppose    prevented   by  the 

strength  of  the  envelope.     At  the  surface 

of  the  ground  (station  n  of  Fig.  531 ;  see 

also  Fig.  526)  let  the  barometer  read  29.6 

inches  and  the  temperature  be  15°  Cent. 

Then  Tn  =  288°  Abs.  Cent.,  and  the  heavi-  FIG.  531. 

ness  of  the  air  at  n  is 

.0807  X  273    2.M  X  14.7 


14.7  288 

273    29.6 

'm-ir-*™1 

At  the  unknown  height  7i,  where  the  balloon  is  to  come  to 
rest,  i.e.,  at  J/~,  Gl  must  =  Vy    [eq.  (2)]  ;  therefore 

Gl  760  Ibs. 

ym  =  y-  =  p^r- — r-sr-  =  -0633  lbs-  Per  cub. 


and  if  the  temperature  at  M  be  estimated  to  be  5°  Cent,  (or 
Tm  —  278°  Abs.  Cent.)  (on  a  calm  day  the  temperature  de- 
creases about  1°  Cent,  for  each  500  ft.  of  ascent),  we  shall 


_.    388 
m        .»      -0633  '  278  ~ 

and  hence,  from  eq.  (5),  §  477,  with  |(7Tm+  Tn)  put  for  TK, 
h  =  26213  X  Iff  X  2.30258  X  log.,,  1.206  =  5088  ft. 


&£ 

CHAPTER  Vt 


HYDRODYNAMICS  BEGUN— STEADY  FLOW  OF  LIQUIDS 
THROUGH  PIPES  AND  ORIFICES. 


489a.  The  subject  of  Water  in  Motion  presents  one  of  the 
most  unsatisfactory  branches  of  Applied  Mechanics,  from  a 
mathematical  stand-point.  The  internal  eddies,  cross-currents, 
and  general  intricacy  of  motion  of  the  particles  among  each 
other,  occurring  in  a  pipe  transmitting  a  fluid,  are  almost  en- 
tirely defiant  of  mathematical  expression,  though  the  flow  of 
water  through  a  circular  orifice  in  a  thin  plate  into  the  air  pre- 
sents a  simpler  case,  where  the  conception  of  "  stream  lines"  is 
probably  quite  close  to  the  truth.  In  most  practical  cases  we 
are  forced  to  adopt  as  a  basis  for  mathematical  investigation 
the  simple  assumption  that  the  particles  move  side  by  side  in 
such  a  way  that  those  which  at  any  instant  form  a  lamina 
or  thin  sheet,  ~|  to  the  axis  of  the  pipe  or  orifice,  remain 
together  as  a  lamina  during  the  further  stages  of  the  flow. 
This  is  the  Hypothesis  of  Flow  in  Plane  Layers,  or  Laminated 
Flow.  Experiment  is  then  relied  on  to  make  good  the  discre- 
pancies between  the  indications  of  the  formulae  resulting  from 
this  theory  and  the  actual  results  of  practice ;  so  that  the  science 
of  Hydrodynamics  is  largely  one  of  coefficients  determined  by 
experiment. "  7 

490.  Experimental  Phenomena  of  a  "  Steady  Flow." — As  pre- 
liminary to  the  analysis  on  which  the  formulae  of  this  chapter 
are  based,  and  to  acquire  familiarity  with  the  quantities  involved, 
it  will  be  advantageous  to  study  the  phenomena  of  the  appara- 
tus represented  in  Fig.  532.  A  large  tank  or  reservoir  I>O  is 
connected  with  another,  DE,  at  a  lower  level,  by  means  of  a 
rigid  pipe  opening  under  the  water-level  in  each  tank.  This 

64G 


STEADY   FLOW    OF  A    LIQUID. 


647 


pipe  lias  no  sharp  curves  or  bends,  is  of  various  sectional  areas 
at  different  parts,  the  changes  of  section  being  very  gradual, 
and  the  highest  point  N^  not  being  more  than  30  ft.  higher 
than  EC)  the  surface-level  of  the  upper  tank.  Let  both  tanks 


FIG.  532. 

be  filled  with  water  (or  other  liquid),  which  will  also  rise  to  H 
and  to  ./Tin  the  pipe.  Stop  the  ends  L  and  N±  of  the  pipe, 
and  through  J/,  a  stop-cock  in  the  highest  curve,  pour  in  water 
to  fill  the  remainder  of  the  pipe ;  then,  closing  M,  unstop  L 
and  NI .  I 

If  the  dimensions  are  not  extreme  (and  subsequent  formulae 
will  furnish  the  means  of  testing/ such  points)  the  water  will 
now  begin  to  flow  from  the  upper  tank  into  the  lower,  and 
all  parts  of  the  pipe  will  continue  full  of  water  as  the  flow 
goes  on. 

Further,  suppose  the  upper  tank  so  large  that  its  surface- 
level  sinks  very  slowly  /  or  that  an  influx  at  A  continually 
makes  good  the  efflux  at  E;  then  the  flow  is  said  to  be  a  Steady 
Flow ;  or,  a  state  of  permanency  is  said  to  exist ;  i.e.,  the  cir- 
cumstances of  the  flow  at  each  section  of  the  pipe  are  per- 
manent, or  steady. 


648  MECHANICS   OF   ENGINEERING. 

By  measuring  the  volume,  V,  of  water  discharged  at  E  in  a 
time  ty  we  obtain  the  volume  of  flow  per  unit  of  time,  viz., 


while  the  weight  of  flow  per  unit  of  time  is 

G  =  QY,  •     •     :     .....     (2) 

where  y  =  heaviness  (§  7)  of  the  liquid  concerned. 

Water  being  incompressible  and  the  pipe  rigid,  it  follows 
that  the  same  volume  of  water  per  unit  of  time  must  be  pass- 
ing at  each  cross-section  of  the  pipe.  But  this  is  equal  to  the 
volume  of  a  prism  of  water  having  F,  the  area  of  the  section^ 
as  a  base,  and,  as  an  altitude,  the  mean  velocity  =  v  with  which 
the  liquid  particles  pass  through  the  section.  Hence  for  any 
section  we  have 

,    .    (3) 

in  which  the  subscripts  refer  to  different  sections.  If  the  flow 
were  unsteady,  e.g.,  if  the  level  BC  were  sinking,  this  would 
be  true  for  a  definite  instant  of  time  ;  but  when  steady,  we 
see  that  it  is  permanently  true;  e.g.,  Flvl  at  any  instant  =  Fjp^ 
at  the  same  or  any  other  instant,  subsequent  or  previous.  In 
other  words,  in  a  steady  flow  the  velocity  at  a  given  section 
remains  unchanged  with  lapse  of  time. 

[N.B.  "We  here  assume  for  simplicity  that  the  different 
particles  of  water  passing  simultaneously  through  a  given  sec- 
tion (i.e.,  abreast  of  each  other)  have  equal  velocities,  viz.,  the 
velocity  which  all  other  particles  will  assume  on  reaching  this 
section.  Strictly,  however,  the  particles  at  the  sides  are  some- 
what retarded  by  friction  on  the  surface  of  the  pipe.  This  as- 
sumption is  called  the  Assumption  of  Parallel  Flow,  or  Flow 
in  Plane  Layers,  or  Laminated  FlowJ] 

Let  us  suppose  Q  to  have  been  found  as  already  prescribed. 
We  may  then,  knowing  the  internal  sectional  areas  at  different 
parts  of  the  pipe,  -ZT,  ,  JVa,  etc.,  compute  the  velocities 


STEADY   FLOW   OF   LIQUIDS.  649 

Vi=Q  +  F>,    v,=  Q  +  F*>    etc., 
«(  a? 

which  the  water  must  have  in  passing  those  sections,  respec- 
tively. It  is  thus  seen  that  the  velocity  at  any  section  has  no 
direct  connection  with  the  height  or  depth  of  the  section  from 
the  plane,  EC,  of  the  upper  reservoir  surface.  The  fraction 

-  will  be  called  the  height  due  to  the  velocity,  v,  or  simply 

*y 

the  velocity-head,  for  convenience. 

Next,  as  to  the  value  of  the  internal  fluid  pressure',  p,  per 
unit-area  (in  the  water  itself  and  against  the  side  or  wall  of 
pipe)  at  different  sections  of  the  pipe.  If  the  end  N±  of  the 
pipe  were  stopped,  the  problem  would  be  one  in  Hydrostatics, 
and  the  pressure  against  the  side  of  the  pipe  at  JVj  (also  at  Nz 
on  same  level)  would  be  simply 


measured  by  a  water  column  of  height 


in  which  pa=  one  atmosphere,  and  Z>  =  34  ft.  =  height  of  an 
ideal  water  barometer,  and  y  =  62.5  Ibs.  per  cubic  foot  ;  and 
this  would  be  shown  experimentally  by  screwing  into  the  side 
of  the  pipe  at  j¥~,  a  small  tube  open/  at  both  ends  ;  the  water 
would  rise  in  it  to  the  level  EC.  That  is,  a  column  of  water 
of  height  —  hl  would  be  sustained  in  it,  which  indicates  that 
the  internal  pressure  at  Nl  corresponds  to  an  ideal  water  col- 
umn of  a  height 


But  when  a  steady  flow  is  proceeding,  the  case  being  now  one 
of  Hydrodynamics,  we  find  the  column  of  water  sustained  at 
rest  in  the  small  tube  (called  an  open  piezometer)  Nfi  has  a 
height  f/1  ,  less  than  7^  ,  and  hence  the  internal  fluid  pressure  is 


650  MECHANICS    OF   ENGINEERING. 

less  than  it  was  when  there  was  no  flow.     This  pressure  being 
called^,  ,  the  ideal  water  column  measuring  it  has  a  height 


at  Nl  ,  and  will  be  called  the  pressure-head  at  the  section  re- 
ferred to.  We  also  find  experteeTfCa11j~iffiat  while  the  flow  is 
steady  the  piezometer-height  y  (and  therefore  the  pressure- 
head  b  -\-  y)  at  any  section  remains  unchanged  with  lapse  of 
time,  as  a  characteristic  of  a  steady  flow. 

[For  correct  indications,  the  extremity  of  the  piezometer 
should  have  its  edges  flush  with  the  inner  face  of  the  pipe 
wall,  where  it  is  inserted.] 

At  ^3,  although  at  the  same  level  as  JV19  we  find,  on  in- 
serting an  open  piezometer,  TF,  that  with  Fz  =  Fl  (and  there- 
fore with  v3  =  v^  yz  is  a  little  less  than  ?/,  ;  while  if  F9  <  Fv 
(so  that  va  >  ^j),  y3  is  not  only  less  than  y,  ,  but  the  dif- 
ference is  greater  than  before.  We  have  therefore  found 
experimentally  that,  in  a  general  way,  when  water  is  flowing 
in  a  pipe  it  presses  less  against  the  side  of  the  pipe  than  it  did 
before  the  flow  was  permitted,  or  (what  amounts  to  the  same 
thing)  the  pressure  between  the  transverse  laminae  is  less  than 
the  hydrostatic  pressure  would  be. 

In  the  portion  HNZ0  of  the  pipe  we  find  the  pressure  less 
than  one  atmosphere,  and  consequently  a  manometer  register- 
ing pressures  from  zero  upward  (and  not  simply  the  excess 
over  one  atmosphere,  like  the  Bourdon  steam-gauge  and  the 
open  piezometer  just  mentioned)  must  be  employed.  At  N^  , 
e.g.,  we  find  the  pressure 

=  J  atmos.,  i.e.,  ^  =  17  ft. 

Even  below  the  level  BC,  by  making  the  sections  quite  nar- 
row (and  consequently  the  velocities  great)  the  pressure  may  be 
made  less  than  one  atmosphere.  At  the  surface  BC  the  pres- 
sure is  of  course  just  one  atmosphere,  while  that  in  the  jet  at 
N^  ,  entering  the  right-hand  tank  under  water,  is  necessarily 

p4  =  1  atmos.  -|-  press,  due  to  col.  hf  of  water  practically  at  rest  ; 


STEADY   FLOW   OF   LIQUIDS.  651 

i.e.,  —  —  pressure-bead  at  J\\  =  I  -f-  A'; 

(whereas  if  N^  were  stopped  by  a  diaphragm,  tbe  pressure- 
bead  just  on  tbe  right  of  tbe  diaphragm  would  be  ~b  -f-  A',  and 
that  on  the  left  £  +  A4.) 

Similarly,  when  a  jet  enters  the  atmosphere  in  parallel  fila- 
ments its  particles  are  under  a  pressure  of  one  atmosphere,  i.e., 
their  pressure-head  =  1)  =  34  ft.  (for  water)  ;  for  the  air  im- 
mediately around  the  jet  may  be  considered  as  a  pipe  between 
which  and  the  water  is  exerted  a  pressure  of  one  atmosphere. 

491.  Recapitulation  and  Examples.  —  We  have  found  experi- 
mentally, then,  that  in  a  steady  flow  of  liquid  through  a  rigid 
pipe  there  is  at  each  section  of  the  pipe  a  definite  velocity  and 
pressure  which  all  the  liquid  particles  assume  on  reaching  that 
section  ;  in  other  words,  at  each  section  of  the  pipe  the  liquid 
velocity  and  pressure  remain  constant  with  progress  of  time. 

EXAMPLE  1.  —  If  in  Fig.  532,  the  flow  having  become  steady, 
the  volume  of  water  flowing  in  3  minutes  is  found  on  meas- 
urement to  be  134  cub.  feet,  the  volume  per  second  is,  from 
<3q.  (1),  §  490, 

Q  —  ||4.  =  0.744  cub.  ft.  per  second. 

EXAMPLE  2.—  If  the  flow  in  2  min.  20  sec.  is  386.4  Ibs.,  the 
volume  of  flow  per  second  is  [ft.,  lb.,  sec.  ;  eqs.  (1)  and  (2)] 

V       G  386.4      1 

Q  —  —  =;  --  r-  1  =  -  .  ---  =  0.0441  cub.  ft.  per  sec. 
t         y  62.5      140 

EXAMPLE  3.  —  In  Fig.  532  the  height  of  the  open  piezometer 
at  Nl  is  yl  =  9  feet  ;  what  is  the  internal  fluid  pressure  ? 
[Use  the  inch,  lb.,  and  sec.]  The  internal  pressure  is 

Pi  =pa  +  y^  —  14.  Y  +  108  X  -&&  =  18.6  Ibs.  per  sq.  inch. 


The  pressure  on  the  outside  of  the  pipe  is,  of  course,  one  at- 
mosphere, so  that  the  resultant  bursting  pressure  at  that  point 

is  3.9  Ibs.  per  sq.  in. 
EXAMPLE  4.  —  The  volume  of  flow  per  second  being  .0441 


652  MECHANICS   OF   ENGINEERING. 

cub.  ft.  per  sec.,  as  in  Example  1,  required  the  velocity  at  a 
section  of  the  (circular)  pipe  where  the  diameter  is  2  inches. 
[Use  ft,  lb.,  and  sec.] 


O       0. 

^2  persec>; 


while  at  another  section  of  the  pipe  where  the  diameter  is  four 
inches  (double  the  former)  and  the  sectional  area,  F,  is  there- 
fore four  times  as  great,  the  velocity  is  \  of  2.02  =  0.505  ft. 
per  sec. 

492.  Bernoulli's  Theorem  for  Steady  Flow;  without  Friction.— 
If  the  pipe  is  comparatively  short,  without  sudden  bends, 
elbows,  or  abrupt  changes  of  cross-section,  the  effect  of  friction 
of  the  liquid  particles  against  the  sides  of  the  pipe  and  against 
each  other  (as  when  eddies  are  produced,  disturbing  the  paral- 
lelism of  flow)  is  small,  and  will  be  neglected  in  the  present 
analysis,  whose  chief  object  is  to  establish  a  formula  for  steady 
flow  through  a  short  pipe  and  through  orifices. 

An  assumption,  now  to  be  made,  oiflow  in  plane  layers,  or 
laminated  flow,  i.e.,  flow  in  laminae  ~|  to  the  axis  of  the  pipe 
at  every  point,  may  be  thus  stated  :  (see  Fig.  533,  which  shows 
a   steady   flow   proceeding,  through   a 
pipe  CD  of  indefinite  extent.)    All  the 
liquid  particles  which    at   any  instant 
form  a  small  lamina,  or  sheet,  as  ABy 
~1  to  axis  of  pipe,  keep  company  as  a 
lamina    throughout    the    whole    flow. 
FIG.  533.  The  thickness,  ds',  of  this  lamina  re- 

mains constant  so  long  as  the  pipe  is  of  constant  cross-section, 
but  shortens  up  (as  at  C)  on  passing  through  a  larger  section, 
and  lengthens  out  (as  at  D)  in  a  part  of  the  pipe  where  the 
section  is  smaller  (i.e.,  the  sectional  area,  F,  is  smaller).  The 
mass  of  such  a  lamina  is  Fds'y  -r-  g  [§  55],  its  velocity  at  any 
section  will  be  called  v  (pertaining  to  that  point  of  the  pipe's 
axis),  the  pressure  of  the  lamina  just  behind  it  is  Fp,  upon  the 
rear  face,  while  the  resistance  (at  the  same  instant)  offered  by 
its  neighbor  just  ahead  is  F(  p  -\-  dp)  on  the  front  face  ;  also 


BERNOULLI  S  'THEOREM. 


653 


its  weight  is  the  vertical  force  Fds'y.     Fig.  534  shows,  as  a 
free  body,  the  lamina  which  at 
any  instant  is  passing  a  point 
A  of  the  pipe's  axis,  where  the 
velocity  is  v  and  pressure  jp. 

Note  well  the  forces  acting ; 
the  pressures  of  the  pipe  wall 
on  the  edges  of  the  lamina  have 
no  components  in  the  direction 
of  v,  for  the  wall  is  considered 
smooth,  i.e.,  those  pressures  are 
1  to  wall;  in  other  words, -no  FIG.  534. 

friction  is  considered.  To  this  free  body  apply  eq.  (7)  of  §  74, 
for  any  instant  of  any  curvilinear  motion  of  a  material  point 

vdv  —  (tang,  acceleration)  X  ds,  .    .     .     .     (1)' 

in  which  ds  =  a  small  portion  of  the  path,  and  is  described  in 
the  time  dt.  Now  the  tang,  accel.  =  J£(tang.  compons.  of  the 
acting  forces)  -f-  mass  of  lamina,  i.e., 


tang.  ace.  = 


«* 


-=-  g 


(2) 


Now,  Fig.  535,  at  a  definite  instant  of  time,  conceive  the 
volume  of  water  in  the  pipe  to  be  subdivided  into  a  great 
number  of  laminae  of  equal  mass  (which  implies  equal  volumes 


FIG.  535. 


in  the  case  of  a  liquid,  but  not  with  gaseous  fluids),  and  let  the 
ds  just  mentioned  for  any  one  lamina  be  the  distance  from  its 
centre  to  that  of  the  one  next  ahead ;  this  mode  of  subdivision 


6,)4  MECHANICS   OF   ENGINEERING. 

makes  the  ds  of  any  one  lamina  identical  in  value  with  its 
thickness  ds'  ,  i.e., 


have  also 

ds  cos  </>  =  —  dz]     or  ds'  cos  0  =  —  da  ;  .     .     (4) 

z  being  the  height  of  the  centre  of  a  lamina  above  any  con- 
venient  horizontal  datum  plane.  Substituting  from  (2),  (3), 
and  (4)  in  (1),  we  derive  finally 

±vdv  +  ^dp  +  dz  =  0.   .  (5V7 

9  > 

The  flow  being  steady,  and  the  subdivision  into  laminae 
being  of  the  nature  just  stated,  each  lamina  in  some  small  time 
dt  moves  into  the  position  which  at  the  beginning  of  dt  was 
filled  by  the  lamina  next  ahead,  and  acquires  the  same  velocity, 
the  same  pressures  on  its  faces,  and  the  same  value  of  z,  that 
the  front  lamina  had  at  the  beginning  of  dt. 

Hence,  considering  the  simultaneous  advance  made  by  all 
the  laminae  in  this  same  dt,  we  may  write  out  an  equation  like 
(5)  for  each  of  the  laminae  between  any  two  cross-sections  n  and 
m  of  the  pipe,  thus  obtaining  an  infinite  number  of  equations, 
from  which  by  adding  corresponding  terms,  i.e.,  by  integra- 
tion, we  obtain 


whence,  performing  the  integrations  and  transposing, 

vm*       Pm\,      _  vn    ,   Pn  ,  ,  (  Bernoulli  s  \  m 

~2g~'~y~     m~"ty     "y~"~n*     '(    Theorem     \   ' 

Denoting  by  Potential  Head  the  vertical  height  of  any  section 
of  the  pipe  above  a  convenient  datum  level,  we  may  state 
Bernoulli's  Theorem  as  follows  : 

In  steady  flow  without  friction,  the  sum  of  the  velocity- 
head,  pressure-head,  and  potential  head  at  any  section  of  the 
pipe  is  a  constant  quantity,  being  equal  to  the  sum  of  the  cor- 
responding heads  at  any  other  section. 


APPLICATIONS   OF   BERNOULLI'S   THEOREM. 


655 


It  is  noticeable  that  in  eq.  (7)  each  of  the  terms  is  a  linear 
quantity,  viz.,  a  height,  or  head,  either  actual,  such  as  zn  and 
zm ,  or  ideal  (all  the  others),  and  does  not  bring  into  account  the 
absolute  size  of  the  pipe,  nor  even  its  relative  dimensions  (vm 
and  vn,  however,  are  connected  by  the  equation  of  continuity 
Fmvm  —  Fnvn\  and  contains  no  reference  to  the  volume  of 
water  flowing  per  unit  of  time  \_Q]  or  the  shape  of  the  pipe's 
axis.  "When  the  pipe  is  of  considerable  length  compared  with 
its  diameter  the  friction  of  the  water  on  the  sides  of  the  pipe 
cannot  be  neglected  (§  512). 

It  must  be  remembered  that  Bernoulli's  Theorem  does  not 
hold  unless  the  flow  is  steady^  i.e.,  unless  each  lamina,  in  com- 
ing into  the  position  just  vacated  by  the  one  next  ahead  (of 
equal  mass),  conies  also  into  the  exact  conditions  of  velocity 
and  pressure  in  which  the  other  was  when  in  that  position. 

[N.B.  This  theorem  can  also  be  proved  by  applying  to  all 
the  water  particles  between  n  and  m,  as  a  collection  of  small 
rigid  bodies  (water  being  incompressible)  the  theorem  of  Work 
and  Energy  for  a  collection  of  Rigid  Bodies  in  §  142,  eq.  (xvi), 
taking  the  respective  paths  which  they  describe  simultaneously 
in  a  single  dt.~\ 

493.  First  Application  of  Bernoulli's  Theorem  without  Friction. 
— Fig.  536  shows  a  large  tank  from  which  a  vertical  pipe  of 
uniform  section  leads  to  another  tank  and  dips  below  the  sur- 
face of  the  water  in  the  latter.  Both  surfaces  are  open  to  the 
air.  The  vessels  and  pipe  being  filled  with 
water,  and  the  lower  end  m  of  the  pipe  un- 
stopped, a  steady  flow  is  established  almost 
immediately,  the  surface  BC  being  very 
large  compared  with  F,  the  area  of  the  (uni- 
form) section  of  the  pipe.  , 

Given  F,  and  the  heights  hQ  and  A,  re- 
quired the  velocity  vm  of  the  jet  at  m  and 
also  the  pressure,  pn,  at  n  (in  pipe  near  en- 
trance of  same),  m  is  in  the  jet,  just  clear 
of  the  pipe,  and  practically  in  the  water- 
level,  AD.  The  velocity  vm  is  unknown, 
but  the  pressure  pm  is  practically  =pa=  one  atmosphere,  since 


FIG.  536. 


656 


MECHANICS   OF  ENGINEERING. 


the  pressure  on  the  sides  of  the  jet  is  necessarily  the  hydro- 
static pressure  due  to  a  slight  depth  below  the  surface  AD. 

.*.  Press.-head  at  m  is  —  =  —  =  I  =  34  feet.     .     .    '(§  423) 

y      y 

Now  apply  Bernoulli's  Theorem  to  sections  m  and  n,  taking 
a  horizontal  plane  through  m  as  a  datum  plane  for  potential 
heads,  so  that  zn  =  h  and  zm  =  0,  and  we  have 


(1) 


But,  assuming  that  the  section  of  the  pipe  is  filled  at  every 
point,  we  must  have 


for,  in  the  equation  of  continuity 

j7  v    __  2?  v 

if  we  put  Fm  =  Fn,  the  pipe  being  of  uniform  section,  we  ob- 


tain  vm  =  vn 


Hence  eq.  (1)  reduces  to 
£2  =  b  -  h  =  34  f t.  -  h. 

y 


(2) 


Hence  the  pressure  at  n  is  less  than  one  atmosphere,  and  if  a 
small  tube  communicating  with  an  air-tight  receiver  full  of  air 
were  screwed  into  a  small  hole  at  n,  the  air  in 
the  receiver  would  gradually  be  drawn  off  until 
its  tension  had  fallen  to  a  value  pn.  [This  is  the 
principle  of  SprengeVs  air-pump,  mercury,  how- 
ever, being  used  instead  of  water,  as  for  this 
heavy  liquid  &  =  only  30  inches.] 

If  h  is  made  >  &  for  water,  i.e.  >  34  feet  (or 
>  30  inches  for  mercury),  pn  would  be  negative 
from  eq.  (2),  which  is  impossible,  showing  that 
the  assumption  of  full  pipe-sections  is  not  borne 
out.  In  this  case,  h  >  &,  only  a  portion,  mri ', 
(in  length  somewhat  less  than  &,)  of  the  tube  will  be  kept  full 


FIG.  537. 


APPLICATIONS    OF   BEENOULLl'S   THEOREM.  657 

during  the  flow  (Fig.  537);  while  in  the  part  AV  vapor  of 
water,  of  low  tension  corresponding  to  the  temperature 
{§  469),  will  surround  an  internal  jet  which  does  not  fill  the 
pipe.  As  for  the  value  of  vm,  Bernoulli's  Theorem,  applied 
to  BC  and  w,  in  Fig.  536,  gives  finally  vm  =  V%gh0 . 

EXAMPLE. — If  h  =  20  feet,  Fig.  536,  and  the  liquid  is  water, 
the  pressure-head  at  n  is  (ft.,  lb.,  sec.) 

£*  =  b  -  h  =  34'  -  20'  =  14  ft., 

r 

and  therefore 
pn=14x  62.5  =  875  Ibs.  per  sq.  ft.  =  6.07  Ibs.  per  sq.  in. 

494.  Second  Application  of  Bernoulli's  Theorem  without  Fric- 
tion.— Knowing  by  actual  measurement  the  open  piezometer 
height  yn  at  the  section  n  in 
Fig.  538  (so  that  the  pressure-  B^-  — 


head,  —  =  I  -f-  yn ,  at  n  is 

r 

known) ;  knowing  also  the 
vertical  distance  hn  from  n 
to  m,  and  the  respective 
cross-sections  Fn  and  ^m  (Fm  being  the  sectional  area  of  the 

jet,  flowing  into  the  air,  so  that  —  =  &),  required  the  volume 
of  flow  per  sec.;  i.e.,  required  Q,  which 

=  Fnvn  =  Fmvm.  (1} 

4  n    n  m    m  \    / 

The  pipe  is  short,  with  smooth  curves, -if  any,  and  friction 
will  therefore  be  neglected.  From  Bernoulli's  Theorem  [eq. 
(7),  §  492J,  taking  m  as  a  datum  plane  for  potential  heads,  we 
have 


But  from  (1)  we  have 


MECHANICS    OF   ENGINEERING. 


substituting  which  in  (2)  we  obtain,  solving  for  vr 


and  hence  the  volume  per  unit  of  time  becomes  known,  viz.> 

Q  =  Fmvm.    ......  '•'.    (4) 


.  —  If  the  cross-section  Fm  of  the  nozzle,  or  jet,  is  >  Fn  / 
vm  becomes  imaginary  (unless  yn  is  negative  (i.e.,  pn  <  one  at- 
mos.),  and  numerically  >  hn)  ;  in  other  words,  the  assigned 
cross-sections  are  not  filled  by  the  flow. 

EXAMPLE.  —  If  yn  —  17  ft.  (thus  showing  the  internal  fluid 
pressure  at  n  to  be  pn  =  y(yn  +  ft)  =  1-J-  atmos.),  An  =  10  ft., 
and  the  (round)  pipe  is  4  inches  in  diameter  at  n  and  3  inches 
at  the  nozzle  m,  we  have  from  (3)  (using  ft.-lb.-sec.  system  of 
units  in  which  g  —  32.2) 


4/2x32.2(17-     L0)_  =  ^ 


[N.B.  Since  Fm  -r-  7^  is  a  ratio  and  therefore  an  abstract 
number,  the  use  of  the  inch  in  the  ratio  will  give  the  same 
result  as  that  of  the  foot.] 

Hence,  from  (4), 

Q  =  Fmvm  =  %7c(-fzJ  X  50.4  =  2.474  cub.  ft.  per  sec. 

495.  Orifices  in  Thin  Plate,— Fig.  539.  When  efflux  takes 
place  through  an  orifice  in  a  thin  plate,  i.e.,  a  sharp-edged 
orifice  in  the  plane  wall  of  a  tank,  a  contracted  vein  (or  "  vena 


ORIFICE   IN  THIN   PLATE. 


659 


contracta")  is  formed,  the  filaments  of  water  not  becoming 
parallel    until    reaching   a    plane,  w, 
parallel  to  the   plane  of  vessel  wall, 
which  for  circular  orifices  is  at  a  dis- 
tance from  the  interior  plane  of  vessel 
wall  equal  .to  the  radius  of  the  circular 
aperture ;  and  not  until  reaching  this 
plane    does   the   internal  fluid   pres- 
sure become  equal  to  that  of  the  sur- 
rounding medium  (atmosphere,  here),  FlG- 539- 
i.e.,  surrounding  the  jet.     We  assume   that  the"  width  of  the 
orifice  is  small   compared   with  A,  unless   the   vessel  wall  is 
horizontal. 

The  area  of  the  cross-section  of  the  jet  at  m,  called  the  con- 
tracted section,  is  found  on  measurement  to  be  from  .60  to  .64 
of  the  area  of  the  aperture  with  most  orifices  of  ordinary 
shapes,  even  with  widely  different  values  of  the  area  of  aper- 
ture and  of  the  height,  or  head,  A,  producing  the  flow.  Call- 
ing this  abstract  number  [.60  to  .64]  the  Coefficient  of  Con- 
traction,  and  denoting  it  by  C,  we  may  write 

Fm=GF, 

in  which.  F  =  area  of  the  orifice,  and  Fm  =  that  of  the  con- 
tracted section.  C  ranges  from  .60  to  .64  with  circular*orifices, 
but  may  have  lower  values  with  some  rectangular  forms.  (See 
table  in  §  503.) 

A  lamina  of  particles  of  water  is  under  atmospheric 
pressure  at  n  (the  free  surface  of  the  water  in  tank  or  reser- 
voir), while  its  velocity  at  n  is  practically  zero,  i.e.  vn  =  0 
(the  surface  at  B  being  very  large  compared  with  the  area  of 
orifice).  It  experiences  increasing  pressure  as  it  slowly  de- 
scends until  in  the  immediate  neighborhood  of  the  orifice, 
when  its  velocity  is  rapidly  accelerated  and  pressure  decreased, 
in  accordance  with  Bernoulli's  Theorem,  and  its  shape  length- 
ened out,  until  finally  at  m  it  forms  a  portion  of  a  filament  of 
a  jet,  its  pressure  is  one  atmosphere,  and  its  velocity,  =  vm , 
we  wish  to  determine.  The  course  of  this  lamina  we  call  a 


660 


MECHANICS    OF    ENGINEERING. 


"  stream-line"  and  Bernoulli's  Theorem  is  applicable  to  it, 
just  as  if  it  were  enclosed  in  a  frictionless  pipe  of  the  same 
form.  Taking  then  a  datum  plane  through  the  centre  of  m, 
we  have 


while 


=  0,     and     i>w=?; 


-       also  =b,    zn  =  h,  .  and    vn  =  0. 


Hence  Bernoulli's  Theorem  gives 


(1) 


and 


That  is,  ^0  velocity  of  the  jet  at  m  is  theoretically  the  same  as 
that  acquired  ~by  a  body  falling  freely  in  vacuo  through  a 
height  =h=  the  "  head  of  water."  We  should  therefore  ex- 
pect that  if  the  jet  were  directly  ver- 
tically upward,  as  at  m,  Fig.  540, 

a  height  j!        _ 

would  be  actually  attained.  [See 
§§  52  and  53.]  Experiment  shows 
that  the  height  of  the  jet  (at  m) 
does  not  materially  differ  from  h  if 
h  is  not  >  6  or  8  feet.  For  h  >  8  ft.,  however,  the  actual  height 
reached  is  <  A,  the  difference  being  not  only  absolutely  but 
relatively  greater  as  h  is  taken  greater,  since  the  resistance  of 
the  air  is  then  more  and  more  effective  in  depressing  and 
breaking  up  the  stream. 

At  m',  Fig.  540,  we  have  a  jet,  under  a  head  =  A',  directed 


FIG.  540. 


OKIFICE   IN   THIN   PLATE. 


at  an  angle  <*0  with  the  horizontal.  Its  form  is  a  parabola 
(§  81),  and  the  theoretical  height  reached  is  h"  =  ti  sina  a0 

(§  80)-  _ 

The  jet  from  an  orifice  in  thin  plate  is  very  limpid  and  clear. 

From  eq.  (1),  we  have  theoretically 

vm= 

(an  equation  we  shall  always  use  for  efflux  into  the  air  through 
orifices  and  short  pipes  in  the  plane  wall  of  a  large  tank  whose 
water-surface  is  very  large  compared  with  the  orifice,  and  is 
open  to  the  air),  but  experiment  shows  that  for  an  "  orifice  in 
thin  plate"  this  value  is  reduced  about  3$  by  friction  at  the 
edges,  so  that  for  ordinary  practical  purposes  we  may  write 

vm  =  (f>V~fyh  =  Q.WV'fyh,     ....    (2) 

in  which  0  is  called  the  coefficient  of  velocity. 

Hence  the  volume  of  flow,  Q,  per  time-unit  will  be 

~  Q  =  Fmvm  =  CF(f>  Vfyh,   on  the  average  =  Q.^FVltyh.   (3)     "  e 


It  is  to  be  understood  that  the  flow  is  steady,  and  that  the 
reservoir  surface  (very  large)  and  the  jet  are  both  under  at- 
mospheric pressure.  0(7  is  called  the  coefficient  of  efffax* 

EXAMPLE  1.  —  Fig.  539.  Required  the  velocity  of  efflux, 
vm  ,  at  m,  and  the  volume  of  the  flow  per  second,  Q,  into  the 
air,  if  h  =  21  ft.  6  inches,  the  circular  orifice  being  2  in.  in 
diam.  ;  take  C  =  0.64.  [Ft.,  lb.,  and  sec.] 

From  eq.  (2), 


vm  =  0.97  1/2  X  32.3  X  21.5  =  36.1  ft.  per  sec. ; 
hence  the  discharge  is 

Q  =  Fmvm  =  0.64  X  fu|j  X  36.1  =  0.504  cub.  ft.  per  second. 

EXAMPLE  2. — [Weisbach.]     Under  a  head  of  3.396  metres 
the  velocity  vm  in  the  contracted  section  is  found  by  measure- 


662  MECHANICS    OF   ENGINEERING. 

ments  of  the  jet-curve  to  be  7.98  metres  per  sec.,  and  the  dis- 
charge proves  to  be  0.01825  cub.  metres  per  sec.  Required 
the  coefficient  of  velocity  (0)  and  that  of  contraction  ((7),  if 
the  area  of  the  orifice  is  36.3  sq.  centimetres. 

Use  the  metre-Jcilogram-second  system  of  units,  in  which 
g  —  9.81  met.  per  sq.  second. 

From  eq.  (2), 


V  2g/i       V2  X  9.81  +  3.396 
while  from  (3)  we  have 

xv  vis  vi^  •v/-LO^O  s^     s*t*.^ 

O= %_  —  =  -~-  = — = —  =  0.631. 

F<t>  V  %y/i      ^vm      ttfiftr  X  7.98 

0  and  (7,  being  abstract  numbers,  are  independent  of  the  sys- 
tem of  concrete  units  adopted. 

NOTE. — To  find  the  velocity  vm  of  the  jet  at  the  orifice  by 
measurements  of  the  jet-curve,  as  mentioned  in  Example  2, 
we  may  proceed  as  follows :  Since  we  cannot  very  readily  as- 
sure ourselves  that  the  direction  of  the  jet  at  the  orifice  is 
horizontal,  we  consider  the  angle  aa  of  the  parabola  (see  Fig. 
93  and  §  80)  as  unknown,  and  therefore  have  two  unknowns 
to  deal  with,  and  obtain  the  necessary  two  equations  by  meas- 
uring the  x  and  y  (see  page  84)  of  two  points  of  the  jet,  re- 
membering that  if  we  use  the  equation  (3)  of  page  84  in  its 
present  form  points  of  the  jet  below  the  orifice  will  have  nega- 
tive 2/'s.  The  substitution  of  these  values  xl ,  o?2 ,  yl ,  and  ?/2 
in  equation  (3)  furnishes  two  equations  between  constants,  in 
which  only  a0  and  A  are  unknown.  To  eliminate  or0 ,  for 

we  write  1  4-  tan3  a0 ,  and  taking  a?2  =  %xl  for  COIL- 
COS  a0 

venience,  we  finally  obtain 

h^-.^-^-    ^V~^      '    aTld  "'*  Vm  ~\/  —  Q~         > 

in  which  yl  and  y2  are  the  vertical  distances  of  the  two  points 


HOUNDED   ORIFICE. 


663 


chosen  "below  the  orifice  ;  that  is,  we  have  already  made  them 
negative  in  eq.  (3)  of  page  84.  The  h  of  the  preceding  equa- 
tion simply  denotes  vm*  -r-  2^,  and  must  not  be  confused  with 
that  of  the  last  two  figures.  For  accuracy  the  second  point 
should  be  as  far  from  the  orifice  along  the  jet  as  possible. 

496,  Orifice  with  Bounded  Approach. — Fig.  541  shows  the 
general  form  and  proportions  of  an  orifice  or  mouth-piece  in 
the  use  of  which  contraction  does  not 
take  place  beyond  the  edges,  the  inner 
surface  being  one  uof  revolution,"  and 
so  shaped  that  the  liquid  filaments  are 
parallel  on  passing  the  outer  edge  m; 
hence  the  pressure-head  at  m  is  =  b 
(=  34  ft.  for  water  and  30  inches  for 
mercury)  in  Bernoulli's  Theorem,  if 
efflux  takes  place  into  the  air.  We 
have  also  the  sectional  area  Fm  —  F  =•  that  of  final  edge  of 
orifice,  i.e.,  the  coefficient  of  contraction,  or  (7,  =  unity  =  1.00, 
so  that  the  discharge  per  time-unit  has  a  volume 

\        ^*1 
Q  _  TT  v  _  jfv 

*&  —  •*  m  um  —  <*•  "m*  \4f  *'      n  ' 

The  tank  being  large,  as  in  Fig.  540,  Bernoulli's  Theorem 
applied  to  m  and  n  will  give,  as  before, 


FlG- 541- 


v= 


as  a  theoretical  result,  while  practically  we  write 


and 


(1) 


As  an  average  0  is  found  to  differ  little  from  0.97  with  this 
orifice,  the  same  value  as  for  an  orifice  in  thin  plate  (§  495). 

497.  Problems  in  Efflux  Solved  by  Applying  Bernoulli's 
Theorem. — In  the  two  preceding  paragraphs  the  pressure- 
heads  at  sections  m  and  n  were  each  =jpa-f-  y  =  height  of 


664  MECHANICS   OF  ENGINEEKINO. 

the  liquid  barometer  =  b ;  but  in  the  following  problems  this 
will  not  be  the  case  necessarily.  However,  efflux  is  to  take 
place  through  a  simple  orifice  in  the  side  of  a  large  reservoir, 
whose  upper  surface  (n)  is  very  large,  so  that  vn  may  be  put 
=  zero. 

Problem  I. — Fig.  542.  What  is  the  velocity  of  efflux,  vm ,  at 
the  orifice  m  (i.e.,  at  the  contracted  sec- 
tion, if  it  is  an  orifice  in  thin  plate) 
of  a  jet  of  water  from  a  steam-boiler,  if 
the  free  surface  at  n  is  at  a  height  =  h 
above  m,  and  the  pressure  of  the  steam 
over  the  water  is  pn ,  the  discharge  tak- 
ing place  into  the  air? 

Applying  Bernoulli's  Theorem  to  sec- 
Fio.648.  tion  m  at  the  orifice  [where  the  pres- 

sure-head is  ~b  and  velocity-head  vm*  -f-  2g  (unknown)]  and  to 
section  n  at  water-surface  (where  velocity-head  =  0  and  pres- 
sure-head =  pn-±  y\  we  have,  taking  m  as  a  datum  for  poten- 
tial heads  so  that  zm  —  0  and  zn  —  k< 


(i>v 


EXAMPLE.  —  Let  the  steam-gauge  read  40  Ibs.  (and  hence 
pn  =  54.  7  Ibs.  per  sq.  inch)  and  A  =  2  ft.  4  in.  ;  required  vm. 

Also  if  J^=  2  sq.  in.,  in  thin  plate,  the  volume  of  discharge 
per  sec.  =  Q  =  ?  For  variety,  use  the  inch-lb.-second  system 
of  units,  in  which  g  —  386.4  inches  per  sq.  second,  while 
I  =  408  inches,  and  the  heaviness  of  water,  y,  —  [62.5  -f-  1728] 
Ibs.  per  cubic  inch.  Hence,  from  eq.  (1), 

vm  =  A  /2  X  386.4F-  n^-j:^  -  408  +  28~]  =  J  935'3    i 
y  L62.5  H-  1728  J      (  per  sec. 


in. 


theoretically  ;  but  practically 


PROBLEMS   OF   EFFLUX — ORIFICES. 


vm  =  935.3  X  0.97  =  907  in.  per  sec., 
so  that  the  rate  of  discharge  (volume)  is 
Q  =  0.64  F<om  =  0.64  X  2  X  907  =  1160.96  cub.  in.  per  sec. 

Problem  II. — Fig.  543.  "With  what  velocity,  vm ,  will  water 
flow  into  the  condenser  C  of  a  steam-engine  where  the  tension 
of  the  vapor  is  pm9  <  one  atmosphere,  if 
h  =  the  head  of  water,  and  the  flow  takes 
place  through  an  orifice  in  thin  plate? 
Taking  position  m  in  the  contracted  section 
where  the  filaments  are  parallel,  and  the 
pressure  therefore  equal  to  that  of  the  sur- 
rounding vapor,  viz.,j?TO,  and  position  n  in 
the  (wide)  free  surface  of  the  water  in  the 
tank,  where  (at  surface)  the  pressure  is  one  FIG.  543. 

atmosphere  [and   /.  ^  =  I  =  34  ft.]  and  velocity  practically 

zero ;  we  have,  applying  Bernoulli's  Theorem  to  n  and  m,  tak- 
ing m  as  a  datum  level  for  potential  heads  (so  that  zn  =  h  and 


-si.  •  •  .  .  a> 


V 


and 


as  theoretical  results.     But  practically  we  must  write 


and 


in  which  JP  =  area  of  orifice  in  thin  plate,  and  C=  coefficient 
of  contraction  —  about  0.62  approximately  [see  §  495]. 


666 


MECHANICS    OF   ENGINEERING. 


EXAMPLE.  —  If  in  the  condenser  there  is  a  "vacuum"  of  2TJ 
inches  (meaning  that  the  tension  of  the  vapor  would  support 
2^-  inches  of  mercury,  in  a  barometer),  so  that 

Pm  =  [|f  X  14.7]  Ibs.  per  sq.  inch,  and  Ji  —  12  feet, 

while  the  orifice  is  £  inch  in  diameter  ;  we  have,  using  the  ft., 
lb.,  and  sec., 


X  32.2 


X 


=  51.1  ft.  per  sec.  v 
(We  might  also  have  written,  for  brevity, 

^  =  [2J-  :  30]  X  34  =  2.833, 

since  the  pressure-head  for  one  atmos.  =  34  feet,  for  water. 
Hence,  for  a  circular  orifice  in  thin  plate,  we  have  the  volume 
discharged  per  unit  of  time, 

Q  =  CFv  =  0.62  X  ?f JsV-X  51.1  =  0.0431  cub.  ft.  per  sec. 

4:  \1^' 

497.  Efflux  through  an  Orifice  in  Terms  of  the  Internal  and 
External  Pressures. — Fig.  544.  Let  efflux  take  place  through 
a  small  orifice  from  the  plane  side  of  a  large  tank,  in  which  at 
the  level  of  the  orifice  the  hydrostatic  pressure  was  =  p'  be- 
fore the  opening  of  the  orifice,  that  of  the  medium  surround- 
ing the  jet  being  =p".  "When  a  steady  flow 
is  established,  after  opening  the  orifice,  the 
pressure  in  the  water  on  a  level  with  the  ori- 
fice will  not  be  materially  changed,  except  in 
t££E.||::f  the  immediate  neighborhood  of  the  orifice  [see 
^Srlj  §  495] ;  hence,  applying  Bernoulli's  Theorem 
•'  to  m  in  the  jet,  where  the  filaments  are  parallel, 
and  a  point  ^,  in  the  body  of  the  liquid  and 
at  the  same  level  as  wi,  and  where  the  particles 


FIG.  544. 


are  practically  at  rest  [i.e.,  vn  =  0]  (hence  not  too  near  the 


FORCE-PUMP. 


667 


orifice),  we  shall  have,  cancelling  out  the  potential  heads  which 
are  equal, 


T.    .     .    .    (I/ 


(In  Fig.  544  p'  would  be  equal  to  pa  -\-  hy.)    Eq.  (1)  is  con- 

veniently applied  to  the  jet  produced  by 

a  force-pump,  supposing,    for   simplicity, 

the  orifice  to  be  in  the  head  of  the  pump- 

cylinder,  as  shown  in  Fig.  545.     Let  the 

thrust   (force)    exerted    along  the    piston- 

rod  be  =  P,  and  the   area  of   the  piston 

be  =  F'  .     Then  the  intensity   of  internal 

pressure    produced  in    the    chamber  AB 

(when  the  piston  moves  uniformly)  is 


P 


FIG.  545. 


while  the  external  pressure  in  the  air  around  the  jet  is  simply 
pa  (one  atmos.). 


Tto (1)' 


(KB.  Of  course,  at  points  near  the  orifice  the  internal 
pressure  is  <  p'\  read  §  495.) 

EXAMPLE. — Let  the  force,  or  thrust,  P,  [due  to  steam-pres- 
sure on  a  piston  not  shown  in  figure,]  be  2000  Ibs.,  and  the 
diameter  of  pump-cylinder  be  d  =  9  inches,  the  liquid  being 
•salt  water  (so  that  y  —  64  Ibs.  per  cubic  foot). 

Then 

Fr  =  **•(•&•)'  =  0.442  sq.  ft., 
and  [ft.,  lb.,  sec.] 


668  MECHANICS    OF    ENGINEERING. 

vm  =  0.97^/2  X  32.2  x  gjj^  -  65.4  ft.  per  sec. 

If  the  orifice  is  well  rounded,  with  a  diameter  of  one  inch, 
the  volume  discharged  per  second  is 


Q  =  Fmvm  =  Fvm  =  »(i)*X  65.4  =  0.353  cub.  ft.  per  sec. 

To  maintain  steadily  this  rate  of  discharge,  the  piston  must 
move  at  the  rate  [veloc.  =  v'~\  of 


v'=Q  +  F'  =  .353  -T-        ~      =  0.806  ft.  per  sec., 

and  the  force  P  must  exert  a,  power  (§  130)  of 

L  =  Pvf  =  2000  X  0.816  =  1632  ft.  Ibs.  per  sec. 
=  about  3  horse-power  (or  3  H.  P.). 

If  the  water  must  be  forced  from  the  cylinder  through  a 
pipe  or  hose  before  passing  out  of  a  nozzle  into  the  air,  the 
velocity  of  efflux  will  be  smaller,  on  account  of  "fluid  fric- 
tion" in  the  hose,  for  the  same  P\  such  a  problem  will  be 
treated  later  [§  513].  Of  course,  in  a  pumping-engine,  by  the 
iise  of  several  pump-cylinders,  and  of  air-chambers,  a  practically 
steady  flow  is  kept  up,  notwithstanding  the  fact  that  the  mo- 
tion of  each  piston  is  not  uniform,  and  must  be  reversed  at  the 
end  of  each  stroke. 

498.  Influence  of  Density  on  the  Velocity  of  Efflux  in  the  Last 
Problem.  —  From  the  equation 


of  the  preceding  paragraph,  where  p"  is  the  external  pressure 
around  the  jet,  and  p'  the  internal  pressure  at  the  same  level 
as  the  orifice  but  well  back  of  it,  where  the  liquid  is  sensibly 


RELATION   OF   DENSITY  TO  VELOCITY  OF  EFFLUX.    669 

at  rest,  we  notice  that  for  the  same  difference  of  pressure 
[p'—prr\  the  velocity  of  efflux  is  inversely  proportional  to  the 
square  root  of  the  heaviness  of  the  liquid.  Hence,  for  the 
same  (pf  —  p"\  mercury  would  flow  out  of  the  orifice  with  a 
velocity  only  0.272  of  that  of  water ;  for 


:5=     /  1      .  272 
:8  "  V  13.5  •    1000' 


Again,  assuming  that  the  equation  holds  good  for  the  flow  of 
gases  (as  it  does  approximately  when^'  does  not  greatly  exceed 
p";  e.g.,  by  6  or  8  per  cent),  the  velocity  of  efflux  of  atmospheric 
air,  when  at  a  heaviness  of  0.807  Ibs.  per  cub.  foot,  would  be 


times  as  great  as  for  water,  with  the  same  p'  —  p".    (See 
§  548,  etc.) 


•rl:ViR- 


499.  Efflux  under  Water.  Simple  Orifice.— Fig.  546.  Let 
and  A2  be  the  depths  of  the  (small)  ori- 
fice below  the  levels  of  the  "  head  "  and 
"  tail "  waters  respectively.  Then,  using 
the  formula  of  §  457,  we  have  for  the 
pressure  at  n  (at  same  level  as  m,  the 
jet) 


FIG.  546. 


and  for  the  external  pressure,  around 
the  jet  at  m, 


whence,  theoretically, 


where  h  =  difference  of  level  between  the  surfaces  of  the  two 
bodies  of  water. 


670 


MECHANICS   OF   ENGINEERING. 


Practically, 


vm  = 


(2) 


but  the  value  of  0  for  efflux  under  water  is  somewhat  uncer- 
tain ;  as  also  that  of  C,  the  coefficient  of  contraction.  Weis- 
bach  says  that  /*,  =  06Y,  is  -^-g-  part  less  than  for  efflux  into  the 
air  ;  others,  that  there  is  no  difference  (Trautwine).  See  also 
p.  389  of  vol.  6,  Jour,  of  Engin.  Associations,  where  it  is 
stated  that  with  a  circular  mouth-piece  of  0.37  in.  diam.,  and 
of  "  nearly  the  form  of  the  vena  contracta"  p  was  found  to  be 
.952  for  discharge  into  the  air,  and  .945  for  submerged  dis- 
charge. 

500.  Efflux  from  a  Small  Orifice  in  a  Vessel  in  Motion. 

CASE  I.  When  the  motion  is  a  vertical  translation  and  uni- 
formly accelerated.  —  Fig.  547.  Suppose  the  vessel  to  move  up- 
ward with  a  constant  acceleration  p. 
(See  §  49a.)  Taking  m  and  n  as  in  the 
two  preceding  paragraphs,  we  know  that 
pm  =p"  =  external  pressure  —  one  at- 


FIG.  547. 


mos.  =pa  (and  /.  &  —  J).     As  to  the 

internal  pressure  at  n  (same  level  as  m, 
but  well  back  of  orifice),  pn  ,  this  is  not 
equal  to  (b  +  h)y,  because  of  the  acceler- 

ated motion,  but  we  may  determine  it  by  considering  free  the 
vertical  column  or  prism  On  of  liquid,  of  cross-section  =  dF, 
the  vertical  forces  acting  on  which  are  padF,  downward  at  0, 
pndF  upward  at  n,  and  its  weight,  downward,  lidFy.  All 
other  pressures  are  horizontal.  For  a  vertical  upward  acceler- 
ation =p,  the  algebraic  sum  of  the  vertical  components  of  all 
the  forces  must  =  mass  X  vert,  accel., 


i.e., 
whence 


(1) 


Putting  pn  and  pa  equal  to  the  p'  and  j9r/,  respectively,  of 
the  equation,  we  have 


ORIFICE  IN   MOTION". 


671 


of  §  497, 


(2) 


It  must  be  remembered  that  vm  is  the  velocity  of  the  jet  rel- 
atively to  the  orifice,  which  is  itself  in  motion  with  a  variable 
velocity.  The  absolute  velocity  wm  of  the  particles  of  the  jet 
is  found  by  the  construction  in  §  83,  being  represented  graph- 
ically by  the  diagonal  of  a  parallelogram  one  of  whose  sides  is 
vm,  and  the  other  the  velocity  c  with  which  the  orifice  itself  is 
moving  at  the  instant,  as  part  of  the  vessel.  The  jet  may 
make  any  angle  with  the  side  of  the  vessel. 

On  account  of  the  flow  the  internal  pressures  of  the  water 
against  the  vessel  are  no  longer  balanced  horizontally,  and  the 
latter  will  swing  out  of  the  vertical  unless  properly  constrained. 

If  p  —  g  =  acc.  of  gravity,  vm  =  V  2  V  2gh.  v  If  p  is  nega- 
tive and  =  g,  vm  —  0 ;  i.e.,  there  is  no  flow,  but  both  the  vessel 
and  its  contents  fall  freely,  without  mutual  action. 

CASE  II.  When  the  liquid  and  the  vessel  both  have  a  uni- 
form rotary  motion  about  a  vertical  axis  with  an  angular  veloc- 
ity =  GO  (§  110).  Orifice  small,  so  that  we  may  consider  the 
liquid  inside  (except  near  the  orifice)  to 
be  in  relative  equilibrium.  Suppose  the 
jet  horizontal  at  m,  Fig.  548,  and  the 
radial  distance  of  the  orifice  from  the 
axis  to  be  =  x.  The  external  pressure 
pm  =  pa ,  and  the  internal  [see  §  428, 

eqs.  (3)  and  (4)]  is 

G0V 


FIG.  548. 

hence  the  velocity  of  the  jet,  relatively  to  the  orifice,  is  (from 
§  497,  since  pn  and  pm  correspond  to  the  pf  and  p"  of  that 
article), 

Apn~Pm}_ 


<*»)', 


672  MECHANICS   OF  ENGINEEKING. 


(3) 


in  which  w,  =  cox,  =  the  (constant)  linear  velocity  of  the  ori- 
fice in  its  circular  path.  The  absolute  velocity  wm  of  the  par- 
ticles in  the  jet  close  to  the  orifice  is  the  diagonal  formed  on 
w  and  vm  (§  83)  /  Hence  by  properly  placing  the  orifice  in  the 
casing,  wm  may  be  made  small  or  large',  and  thus  the  kinetic 
energy  carried  away  in  the  effluent  water  be  regulated,  within 
certain  limits.  Equation  (3)  will  be  utilized  subsequently  in 
the  theory  of  Barker's  Mill. 

EXAMPLE.  —  Let  the  casing  make  100  revol.  per  min.  (whence 
G?  =  [2?rlOO  -r-  60]  radians  per  sec.),  A0  =  12  feet,  and  x  =  2 
ft.  ;  then  (ft.,  lb.,  sec.) 


Vm  =       %  X  32.2  X  12  +  =  34.8  ft  per  sec. 


(while,  if  the  casing  is  not  revolving,  vm  =  V2gh0  =  only  27.  8 
ft.  per  sec.). 

If  the  jet  is  now  directed  horizontally  and  backward,  and 
also  tangentially  to  the  circular  path  of  the  centre  of  the  orifice, 
its  absolute  velocity  (i.e.,  relatively  to  the  earth)  is 

wm  =  vm  —  cox=  34.8  —  20.9  =  13.9  ft.  per  sec., 

and  is  also  horizontal  and  backwards.  If  the  volume  of  flow 
is  Q  =  0.25  cub.  feet  per  sec.,  the  kinetic  energy  earned  away 
with  the  water  per  second  (§  133)  is 


ft.  Ibs.  per  second  =  0.085  horse-power. 

501.  Theoretical  Efflux  through  Rectangular  Orifices  of  Con- 
siderable Vertical  Depth,  in  a  Vertical  Plate.  —  If  the  orifice  is 
so  large  vertically  that  the  velocities  of  the  different  filaments 
in  a  vertical  plane  of  the  stream  are  theoretically  different,  hav- 
ing different  "  heads  of  water,"  we  proceed  as  follows,  taking 
into  account,  also,  the  velocity  of  approach,  c,  or  mean  velocity 


EECTANGULAS  OKIFICES. 


673 


(if  any  appreciable),  of  the  water  in  the  channel  approaching 
the  orifice. 

Fig.  549  gives  a  section  of  the  side  of  the  tank  and  orifice. 
Let  b  =  width  of  the  rectangle,  the  sills  of  the  latter  being 
horizontal^  and  a  =  A2  —  A, ,  its  height.  Disregarding  con- 
traction for  the  present,  the  theoretical  volume  of  discharge 
per  unit  of  time  is  equal  to  the 
sum  of  the  volumes  like  v^dF 


(—  vjbdx),  in  which  vm  =  the 
velocity  of  any  filament,  as  m, 
in  the  jet,  and  bdx  =  cross-sec- 
tion of  the  small  prism  which 
passes  through  any  horizontal 
strip  of  the  area  of  orifice,  in  a 
unit  of  time,  its  altitude  being 
vm .  For  each  strip  there  is  a  FIG.  549. 

different  x  or  "  head  of  water,"  and  hence  a  different  velocity. 
Now  the  theoretical  discharge  (volume)  per  unit  of  time  is 

/>*=/-  • 
Q  =  sum  of  the  volumes  of  the  elem.  prisms  =J  vmc 


*dx. 


But  from  Bernoulli's  Theorem,  if  "k  —  c*  -^-  2g  =  the  velocity- 
head  at  7i,  the  surface  of  the  channel  of  approach  nC,b  being 
the  pressure-head  of  n,  and  x  its  potential  head  referred  to  m  as 
datum  (!N~.B.  This  5  =  34  ft.  for  water,  and  must  not  be  con- 
fused with  the  width  5  of  orifice),  we  have  [see  §  492,  eq.  (7)] 


(2)' 

and  since  dx  =  d(x  +  Tc\  Jc  being  a  constant,  we  have,  from  (1)' 
and  (2)', 

Theoret.  O  =  bV~ 


674  MECHANICS   OF   ENGINEERING. 

or 

Theoret.  Q  =  &VTg  [(A,  +  ty  -  &  +  *)»]•      -    (I/ 

(b  now  denotes  the  width  of  orifice.)     If  c  is  small,  the  chan- 
nel of  approach  being  large,  we  have 

Theoret.  Q  =  $>  Vty  (hf —hf)    ......     (2)*' 

(c  being  =  Q  -f-  area  of  section  of  nC).v 

If  Ax  =  0,  i.e.,  if  the  orifice  becomes  a  notch  in  the  side,  or 
an  overfall  [s.ee  Fig.  550,  which  shows  the  contraction  which 
actually  occurs  in  all  these  cases],  we  have  for  an  overfall 


Theoret.  Q  =  &¥  2g\_(h,  +  &)1  -  #J (3) 

NOTE. — Both  in  (1)  and  (2)  h,  and  A2  are  the  vertical  depths. 
...  .  ..  •  • .  .-.•  -.v  •  •  .       °^  the  respective  sills  of  the  orifice 

from  the  surface  of  the  water 
three  or  four  feet  ~back  of  the  plane 
of  the  orifice,  where  the  surface  is 
comparatively  level.  This  must 
be  specially  attended  to  in  deriv- 


Fl°-  55°-  ing  the  actual  discharge  from  the 

theoretical  (see  §  503). 

If  Q  were  the  unknown  quantity  in  eqs.  (1)  and  (3)  it  would 
be  necessary  to  proceed  by  successive  assumptions  and  ap- 
proximations, since  Q  is  really  involved  in  &  ;  for 

and  ~ 


(where  F^  is  the  sectional  area  of  the  channel  of  approach  nC\ 
With  Tc  =  0  (or  c  very  small,  i.e.,  Fn  very  large),  eq.  (3)  re- 
duces (for  an  overfall)  to 


Theoret.  Q  =  |5A2  Vfyh, ,        ......    (3 J) ' 

or  •$  as  much  as  if  all  parts  of  the  orifice  had  the  same  head  of 
water  =  A2  (as  for  instance  if  the  orifice  were  in  the  horizontal 
bottom  of  a  tank  in  which  the  water  was  A2  deep,  the  orifice 
having  a  width  =  b  and  length  =  Aa). 


TRIANGULAR   ORIFICE. 


675 


502,  Theoretical  Efflux  through  a  Triangular  Orifice  in  a  Thin 
Vertical  Plate  or  Wall.  Base  Horizontal.— Fig.  551.  Let  the 
channel  of  approach  be  so  large  that  the  velocity  of  approach 
may  be  neglected,  h^  and  A2  =  depths  of  sill  and  vertex, 
which  is  downward.  The  analysis-  differs  from  that  of  the 
preceding  article  only  in  having  k  =  0  and  the  length  u,  of  a 
horizontal  strip  of  the  orifice,  variable ;  5  being  the  length  of 
the  base  of  the  triangle.  From  similar  triangles  we  have 


—  x 


-(A,-*). 


/.  Theoret.  Q  =  fvmdF=fvmudx= 


and  finally,  substituting  from  eq.  (2)'  of  §  501,  with  Jc  =  0, 


FIG.  551. 


Theoret.  Q  = 


FIG.  552. 


For  a  triangular  notch  as  in  Fig.  552,  this  reduces  to 


Thewet.      = 


15  2 


(5)' 


i.e.,  ^  of  the  volume  that  would  be  discharged  per  unit  of 


676 


MECHANICS    OF   ENGINEERING. 


time  if  the  triangular  orifice  witli  base  b  and  altitude  A2  were 
cut  in  the  horizontal  bottom  of  a  tank  under  a  head  of  ht 
The  measurements  of  ^2  and  b  are  made  with  reference  to  the 
level  surface  back  of  the  orifice  (see  figure) ;  for  the  water- 
surface  in  the  plane  of  the  orifice  is  curved  below  the  level 
surface  in  the  tank. 

Prof.  Thomson  has  found  by  experiment  that  with 
1)  =  2A2,  the  actual  discharge  —  theoret.  disch.  X  0.595  ;  and 
with  I  =  4A2 ,  actual  =  theoret.  disch.  X  0.620. 

503.  Actual  Discharge  through  Sharp-edged  Rectangular  Ori- 
fices (sills  horizontal)  in  the  vertical  side  of  a  tank  or  reservoir. 
CASE  I.   Complete   and    Perfect  Contraction. — The    actual 
volume  of  water  discharged  per  unit  of  time  is  much  less  than 
the  theoretical  values  derived  in  §  501, 
chiefly  on  account  of  contraction.     By 
complete  contraction  we  mean  that  no 
edge  of  the  orifice  is  flush  with  the 
side  or  bottom  of  the  reservoir ;  and 
by  perfect  contraction,  that  the  channel 
of  approach,    to    whose    surface    the 
heads  hl  and  7*2  are    measured,  is  so 
large  that  the  contraction  is  practically 
the  same  if  the  channel  were  of  infi- 
nite extent  sideways  and  •  downward 
from  the  orifice. 

For  this  case  (A,  not  zero)  it  is  found  most  convenient  to 
use  the  following  practical  formula  (b  =  width) : 


Actual  Q  =  / 


+ 


in  which  (see  Fig.  553)  a  =  the  height  of  orifice,  hl  =  the  ver- 
tical depth  of  the  upper  edge  of  the  orifice  below  the  level  of 
the  reservoir  surface,  measured  a  few  feet  back  of  the  plane  of 
the  orifice,  and  /*0  is  a  coefficient  of  efflux  (an  abstract  number), 
dependent  on  experiment. 

With  /i0  =  0.62  approximate  results  (within  3  or  4  per  cent) 
may  be  obtained  from  eq.  (6)  with  openings  not  more  than 


(IY        « 

il    ,  ( 

.  << 


u^«^i  s   \r 

RECTANGULAR  ORIFICES.  677 

18  inches,  or  less  than  1  inch,  high ;  and  not  less  than  1  inch 
wide;  with  heads  (h,  +  -\  from  1  ft.  to  20  or  30  feet. 

EXAMPLE. — What  is  the  actual  discharge  (volume)  per  min- 
ute through  the  orifice  in  Fig.  553,  14  inches  wide  and  1 
foot  high,  the  upper  sill  being  8  ft.  6  in.  below  the  surface  of 
still  water  ?  Use  eq.  (6)  with  the  ft.,  lb.,  and  sec.  as  units,  and 
/*„  =  0.62. 

Solution : 
Q  =  0.62  X  1 X  H  X  V2  X  32.2[8J+^]m  17.41  cub. ft. per. sec. 

while  the  flow  of  weight  is 

G  =  Qy  =  17.41  X  62.5  =  1088  Ibs.  per  second. 

Poncelet  and  Lesbros*  Experiments. — For  comparatively  ac- 
curate results,  values  of  //0  taken  from  the  following  table 
(computed  from  the  careful  experiments  of  Poncelet  and  Les- 
bros)  may  be  used  for  the  sizes  there  given,  and,  where  prac- 
ticable, for  other  sizes  by  interpolation.  To  use  the  table,  the 
values  of  At ,  #,  and  b  must  be  reduced  to  metres,  which  can  be 
done  by  the  reduction-table  below ;  but  in  substituting  in  eq. 
(6),  if  the  metre-kilogram-second  system  of  units  be  used  g 
must  be  put  =  9.81  metres  per  square  second  (see  §  51),  and  Q 
will  be  obtained  in  cubic  metres  per  second. 

Since  /*0  is  an  abstract  number,  once  obtained  as  indicated 
above,  it  does  not  necessitate  any  particular  system  of  units  in 
making  substitutions  in  eq.  (6).  The  ft.,  lb.,  and  sec.  will  be 
used  in  subsequent  examples. 

TABLE  FOR  REDUCING  FEET  AND  INCHES  TO  METRES. 


1 

foot 

=  0.30479 

metre. 

1 

inch 

— 

0.0253 

metre. 

2 

feet 

=  0.60959 

u 

2 

inches 

3= 

0.0507 

u 

3 

u 

=  0.91438 

u 

3 

u 



0.0761 

it 

4 

a 

=  1.21918 

metres. 

4 

u 



0.1015 

tt 

5 

u 

=  1.52397 

u 

5 

u 



0.1268 

tt 

6 

a 

=  1.82877 

u 

6 

u 

— 

0.1522 

tt 

7 

a 

=  2.13356 

tt 

7 

« 



0.1776 

tt 

8 

u 

=  2.43836 

u 

8 

tt 

— 

0.2030 

tt 

9 

u 

=  2.74315 

a 

9 

u 



0.2283 

tt 

10 

u 

=  3.04794 

a 

10 

u 



0.2536 

tt 

11 

u 

— 

0.2790 

tt 

678  MECHANICS   OF  ENGINEERING. 

TABLE,  FROM  PONCELET  AND  LESBROS. 

VALUES  OF  //<>,  FOR  EQ.  (6),  FOR  RECTANGULAR  ORIFICES  IN  THIN  PLATE. 
(Complete  and  perfect  contraction.) 


Value  of  HH 
Fig.  553  (in 

metres). 

6  =  .20*. 
a  =  .20*- 

b  =  .20m- 
a  =  .10»>- 

6  =  .20"" 

a  =  .05ra- 

b  =  .20">- 

a  =  .03m- 

b  =  .20°»- 
a  =  .fej- 

6  =  .20»- 
a  =  .Olm- 

b  =  .80» 
a  =.20" 

6  =  .60m- 
a  =   02m- 

Mo 

Mo 

Mo 

Mo 

Mo 

Mo 

Mo 

Mo 

0.005 

0.705 

.010 

0.607 

0.630 

0.660 

.701 

0.644 

.015 

0.593 

.612 

.632 

.660 

.697 

.644 

.020 

0.572 

.596 

.615 

.634       .659 

.694 

.043 

.030 

.578 

.600 

.620 

.638     '  .659 

.688 

0.593 

.642 

.040 

.582 

.603 

.623 

.640 

.658 

.683 

.595 

.642 

.050 

.585 

.605 

.625 

.640 

.658 

.679 

.597 

.641 

.060 

.587 

.607 

.627 

.640 

.657 

.676 

.599 

.641 

.070 

.588 

.609 

.628 

.639 

.656 

.673 

.600 

.640 

.080 

.589 

.610 

.629 

.638  !     .656 

.670 

.601 

.640 

.090 

.591 

.610 

.629 

.637 

.655 

.668 

.601 

.639 

.100 

.592 

.611 

.630 

.637 

.654 

.666 

.602 

.639 

.120 

.593 

.612 

.630 

.636 

.653 

.663 

.603 

.638 

.140 

.595 

.613 

.630 

.635 

.651 

.660 

.603 

.637 

I  .160 

.596 

.614 

.631 

.634 

.650 

.658 

.604 

.637 

.180 

.597 

.615 

.630 

.634 

.649 

.657 

.605 

.636 

.200 

.598 

.615 

.630 

.633 

.648 

.655 

.605 

.635 

.250 

.599 

.616 

.630 

.632 

.646 

.653 

.606 

.634 

.300 

.600 

.616 

.629 

.632        .644 

.650 

.607 

.633 

.400 

.602 

.617 

.628 

.631 

.642 

.647 

.607 

.631 

.500 

.603 

.617 

.628 

.630 

.640 

.644 

.607 

.630 

.600 

.604 

.617 

.627 

.630 

.638 

.642 

.607 

.629 

.700 

.604 

.616 

.627 

.629 

.637 

.640 

.607 

.628 

.800 

.605 

.616 

.627 

.629 

.636 

.637 

.606 

.62$ 

.900 

.605 

.615 

.626 

.628 

.634 

.635 

.606 

.627 

1.000 

.605 

.615 

.626 

.628 

.633 

.632 

.605 

.626 

1.100 

.604 

.614 

.625 

.627 

.631 

.629 

.604 

.626 

1.200 

.604 

.614 

.624 

.626 

.628 

.626 

.604 

.625 

1.300 

.603 

.613 

.622 

.624 

.625 

.622 

.603 

.624 

1.400 

.603 

.612 

.621 

.622 

.622 

.618 

.603 

.624 

1.500 

.602       .611 

.620 

.620 

.619 

.615 

.602 

.623 

1.600 

.602 

.611 

.618 

.618 

.617 

.613 

.602 

.623 

1.700 

.602 

.610 

.617 

.616 

.615 

.612 

.602 

.622 

1.800         .601        .609 

.615 

.615 

.614 

.612 

.602 

.621 

1.900         .601        .608 

.614 

.013 

.612 

.611 

.602 

.621 

2.000         .601        .607 

.613 

.612 

.612 

.611 

.602 

.620 

3.000         .601        .603 

.606 

.608 

.610 

.609 

.601 

.615 

i              i 

EXAMPLE.  —  With  h,  =  4  in.  [=0.10  met.],  .a  =  8  in. 
[=  0.20  met] ,  Z>  =  1  f t.  8  in.  [=  0.51  met]  ,  required  the 
(actual)  volume  discharged  per  second.  See  Fig.  553. 


RECTANGULAR   ORIFICES. 


679 


From  the  foregoing  table, 

for  h,  =  0.10m-,  I  =  0.60m-  and  a  =  0.20m-,  we  find  /*0  =  .602 
"  h,  =  0.10™-,  I  =  0.20m-    "    a  =  0.20m-,       "        /*.  =  .592 

diff.  =  ."ulO 

Hence,  by  interpolation, 

for  h,  =  0.10m-  J  =  0.51m-,  and  a  =  0.20m-,  we  have 
V0  =  0.602  -  A  [°-602  -  0.592]  =  0.600.v 

Hence  [ft.,  lb.,  sec.],  remembering  that  /*0  is  an  abstract  num- 
ber, from  eq.  (6), 


Q  =  0.600  X 


X 


-  4.36 


cub.  ft.  per  second. 

CASE  II.  Incomplete  Contraction.  —  This  name  is  given  to 
the  cases,  like  those  shown  in  Fig.  554,  where  one  or  more 
sides  of  the  orifice  have  an  interior  border  flush  with  the  sides 
or  bottom  of  the  (square-cornered)  tank. 

Not  only  is  the  general  direction  of  the  stream  altered,  but 
the  discharge  is  greater,  on  account  of  the  larger  size  of  the 
contracted  section,  since  contraction  is  prevented  on  those  sides 
which  have  a  border.  It  is  assumed  that  the  contraction  which 
does  occur  (on  the  other  edges)  is  perfect  /  i.e.,  the  cross-sec- 
tion of  the  tank  is  large,  compared  with  the  orifice.  According 
to  the  experiments  of  Bidone  and 
Weisbach  with  Poncelet's  ori- 
fices (i.e.,  orifices  in  thin  plate 
mentioned  in  the  preceding  table), 
the  actual  volume  discharged  per 
unit  of  time  is 


(7) 


FIG.  554. 


(differing  from  eq.  (6)  only  in 
the  coefficient  of  efflux  /*),  in  which  the  abstract  number  >w  is 
found  thus :  Determine  a  coefficient  of  efflux  ja0  as  if  eq.  (6) 
were  to  be  used  in  Case  I ;  i.e.,  as  if  contraction  were  complete 
and  perfect ;  then  write 


680 


MECHANICS   OF   ENGINEERING, 


/£  =  /!„[!  + 0.155 »], (7)' 

where  n  =  the  ratio  of  the  length  of  periphery  of  the  orifice 
with  a  border  to  the  whole  periphery. 
E.g.,  if  the  lower  sill,  only,  has  a  border, 


while  if  the  lower  sill  and  both  sides  have  a  border, 
w  =  (20  +  &)-*-  [2(0  +  &)]. 

EXAMPLE.—  If  h,  =  8  ft.  (=  2.43m-),  I  =  2  ft.  (=  0.60m-), 
a  =  4  in.  (=  0.10m-),  and  one  side  is  even  with  the  side  of 
the  tank,  and  the  lower  sill  even  with  the  bottom,  required  the 
volume  discharged  per  second.  (Sharp-edged  orifice,  in  ver- 
tical plane,  etc.) 

Here  for  complete  and  perfect  contraction  we  have,  from 
Poncelet's  tables  (Case  I),  /i0  =  0.608.  Now  n  =  i;  hence,, 
from  eq.  (7)', 


=  0.608  [1  +  0.155  X  i]=  0.6551  ; 


hence,  eq.  (7), 


Q  =  0.655  X  2  X  A  V2x32.2(8+i.A) 
=  10.23  cub.  ft.  per  sec. 

CASE  III.  Imperfect  Contraction.  —  If  there  is  a  submerged 
channel  of  approach,  symmetrically 
placed  as  regards  the  orifice,  and  of 
an  area  (cross-section),  =  Gr9  not 
much  larger  than  that,  =  F,  of  the 
orifice  (see  Fig.  555),  the  contraction 
is  less  than  in  Case  I,  and  is  called 
imperfect  contraction.  Upon  his 
experiments  with  Poncelet's  orifices, 
with  imperfect  contraction,  Weisbach. 

bases  the  following  formula  for  the  discharge  (volume)  per 

unit  of  time,  viz., 


(8) 


Fl°-  555- 


EECTANGULAK   OKIFICES. 


681 


(see  Fig.  553  for  notation),  with  the  understanding  that  the  co- 
efficient 

......    (8)' 


where  /<0  IB  the  coefficient  obtained  from  the  tables  of  Case  I 
(as  if  the  contraction  were  perfect  and  complete),  and  ft  an  ab- 
stract number  depending  on  the  ratio  F  :  G  =  ra,  as  follows  : 


=  0.0760  [9™--  1.00]. 


TABLE  A. 


To  shorten  computation  Weisbach  gives  the  following  table 
for/?: 

EXAMPLE.— Let  h,  =  V  9J-"  (=  1.46 
met.),  the  dimensions  of  the  orifice 
being — 

width   =  b  =  8  in.  (=  0.20m-); 
height  =  a  =  5  in.  (=  0.126m-) ; 

while  the  channel  of  approach  (CD, 
Fig.  555)  is  one  foot  square.  From 
Case  I,  we  have,  for  the  given  di- 
mensions and  head, 


m. 

/3. 

m. 

0. 

.05 

.009 

.55 

.178 

.10 

.019 

.60 

.208 

.15 

.030 

.65 

.241 

.20 

.042 

.70 

.278 

.25 

.056 

.75 

.319 

.30 

.071 

.80 

.365 

.35 

.088 

.85 

.416 

.40 

.107 

.90 

.473 

.45 

.128 

.95 

.537 

.50 

.152 

1.00 

.608 

=  0.610; 


We  find  [Table  A] 


144  sq.  in. 


p  =  0.062 ; 


and  hence  ;*  =  /*„  (1.062),  from  eq.  (8)'.     Therefore,  from  eq. 
(8),  with  ft.,  lb.,  and  sec., 


Q  =  0.610  X  1.062  X  A  •  A  V2  X  32.2  X  5 
=  3.22  cub.  ft.  per  sec. 

CASE  IV.  Head  measured  in  Moving  Water. — See  Fig. 
556.  If  the  head  h^ ,  of  the  upper  sill,  cannot  be  measured  to 
the  level  of  still  water,  but  must  be  taken  to  the  surface  of  a 
channel  of  approach,  where  the  velocity  of  approach  is  quite 


682 


MECHANICS   OF  ENGINEERING. 


appreciable,  not  only  is  the  contraction  imperfect,  but 
strictly  we  should  use  eq.  (1)  of  §  501,  in 
which  the  velocity  of  approach  is  considered. 
Let  F  =  area  of  orifice,  and  G  that  of  the 
cross-section  of  the  channel  of  approach ; 
then  the  velocity  of  approach  is  c  =  Q  -f-  G, 
and  ~k  (of  above  eq.)  =  c2  -f-  2^  =  Q'  -f-  2^£2; 
but  Q  itself  being  unknown,  a  substitution  of 
~k  in  terms  of  Q  in  eq.  (1),  §  501,  leads  to  an 
equation  of  high  degree  with  respect  to  Q. 

Practically,  therefore,  it  is  better  to  write 


•    (9) 


and  determine  fit  by  experiment  for  different  values  of  the 
ratio  F  -±-  G.    Accordingly,  "Weisbach  found,  for  Poncelet's 
orifices,  that  if  j*0  is  the  coefficient  for  complete  and  perfect 
contraction  from  Case  I,  we  have 

l«  =  /i.(l  +  /8')  .......     (9)' 

ft'  being  an  abstract  number,  and  being  thus  related  tof-T-  G, 


=  0.641 


(9)" 


h^  was  measured  to  the  surface  one  metre  back  of  the  plane  of 
the  orifice,  and  F  :  G  did  not  exceed  0.50. 

Weisbach  gives  the  following  table  computed  from  eq.  (9)"  : 


TABLE  B. 


EXAMPLE.  —  A  rectangular  water-trough  4  ft. 
wide  is  dammed  up  with  a  vertical  board  in 
which  is  a  rectangular  orifice,  as  in  Fig.  556,  of 
width  b  =  2  ft.  (=  0.60  met.),  and  height  a  =  6 
in.  (=  0.15  met.)  ;  and  when  the  water-level  be- 
hind the  board  has  ceased  rising  (i.e.,  when  the 
flow  has  become  steady),  we  find  that  h1  =  2  ft., 
and  the  depth  behind  in  the  trough  to  be  3  ft. 
Kequired  Q. 

Since  F:  G  =  l  sq.  ft.  -r-  12  sq.  ft.  =  .0833, 
we  find  (Table  B)  ft'  —  0.005  ;  and  ^0  being  =  0.612  from  Pon- 
celet's  tables,  Case  I,  we  have  finally,  from  eq.  (9), 


F-t-G. 

/3'. 

0.05 

.002 

.10 

.006 

.15 

.014 

.20 

.026 

.25 

.040 

.30 

.058 

.35 

.079 

.40 

.103 

.45 

.130 

.50 

.160 

DISCHARGE    OF   OVERFALL-WEIRS. 


683 


Q  =  0.612(1.005)  2  xi  V2  X  32.2  X  2.25 
=  7.41  cub.  f-t.  per  second. 

504.  Actual  Discharge    of   Sharp-edged   Overfalls   (Overfall- 
weirs;  or  Rectangular  Notches  in  a  Thin  Vertical  Plate), 

CASE  I.  Complete  and  Perfect  Contraction  (the  normal 
case),  Fig.  557 ;  i.e.,  no  edge  is  flush 
with  the  side  or  bottom  of  the 
reservoir,  whose  sectional  area  is 
very  large  compared  with  that,  &A3 , 
of  the  notch.  By  depth,  A2 ,  of  the 
notch,  we  are  to  understand  the 
depth  of  the  sill  Mow  the  surface 
a  few  feet  back  of  the  notch  where 
it  is  level.  In  the  plane  of  the 
notch  the  vertical  thickness  of  the  stream  is  only  from  J  to  -^j- 
of  A2 .  Putting,  therefore,  the  velocity  of  approach  =  zero, 
and  hence  k  =  0,  in  eq.  (3)  of  §  501,  we  have  for  the 


FIG.  557. 


Actual  Q  =  ^0|JA2 1/2</A2,  .     .     .  ...    (10) 

(b  =  width  of  notch,)  where  //„  is  a  coefficient  of  efflux  to  be 
determined  by  experiment. 

Experiments  with  overfalls  do  not  agree  as  well  as  might  be 
desired.  Those  of  Poncelet  and  Lesbros  gave  the  results  in 
Table  C. 

EXAMPLE  1. — With 

A2  =  1  ft.  4  in.  (=  .405m-), 
I  —  2  ft.  (=  0.60m-), 

we  have,  from  Table  C,  /*0  =  .586, 
and  (ft.,  lb.,  sec.) 


TABLE  C. 


4/2X32.2XI 
—  9.54  cub.  ft.  per  sec. 

EXAMPLE   2. —  What    width,  5, 
must  be  given  to  a  rectangular  notch,  for  which  7i2  =  10  in. 
(=  0.25m-),  that  the  discharge  may  be  Q  =  6  enb.  feet  per  sec.? 


For  6  = 

O.SO""- 

For  6- 

=  0.60n». 

metres. 

metres. 

ft* 

.01 

.636 

7i3 
.06 

.6% 

.02 

.620 

.08 

.613 

.03 

.618 

.10 

.609 

.04 

.610 

.02 

.605 

.06 

.601 

.15 

.600 

.08 

.595 

.20 

.592 

.10 

.592 

.30 

.586 

.15 

.589 

.40 

.586 

.20 

.585 

.50 

.586 

.22 

.577 

.60 

.585 

For  approx.  results  MO 

=  .60 

C84 


MECHANICS   OF   ENGINEEKING. 


Since  b  is  unknown,  we  cannot  use  the  table  immediately,, 
but  take  //0  =  .600  for  a  first  approximation  ;  whence,  eq.  (10), 
(ft.,  lb.,  sec.,) 


o  = 


6 


0.6  X  1  X  if  t/2  X  32.2  X 


=  2.46  ft. 


Then,  since   this   width   does   not    much  exceed  0.60  metre, 
we  may  take,  in  Table  C,  for  Aa  =  0.25  met.,  /*0  =  .589  ; 


I  = 


6 


.589  X  I  X 


X  32.2  X 


=  2.50  ft. 


CASE  II.  Incomplete  Contraction'  i.e.,  both  ends  are  flush 
with  the  sides  of  the  tank,  these  being  ~\  to  the  plane  of  the 
notch.  According  to  "Weisbach,  we  may  write 

.    ,    .    .  iH*    (11) 


in  which  /£  =  1.041/«0  ,  /i0  being  obtained  from  Table  C  for  the 
normal  case,  i.e.,  Case  I.     The  section  of  channel  of  approach 
is  large  compared  with  that  of  the  notch  ;  if  not,  see  Case  IV. 
CASE  III.  Imperfect  Contraction  •  i.e.,  the  velocity  of  ap- 
proach is  appreciable  ;  the  sectional  area  G 
of  the  channel  of  approach  not  being  much 
larger  than  that,  F,  =  Ma  =  area  of  notch. 
Fig.  558.     b  =  width,   and    Aa  =  depth   of 
notch  (see  Case  I).     Here,  instead  of  using 
a  formula  involving 

k  =  c2  -T-  2^  =  [Q  4-  ay  4-  2^ 

FIG.  558. 

(see  eq.  (3),  §  501),  it  is  more  convenient  to  put 


as  before,  with 


(12) 


(12)' 


in  which  ^0  is  for  the  normal  case  [Case  I]  ;  and  /?,  according 


OVERFALL-WEIK3. 


685 


to  Weisbach's  experiments,  may  be  obtained  from  the  empiri- 
cal formula 


(12)' 


[Table  D  is  computed  from  (12)".] 

(The  contraction  is  complete  in  this  case  ; 
not  flush  with  the  sides  of  the  tank.) 

EXAMPLE.  —  If  the  water  in  the  channel  of  ap- 
proach has  a  vertical  transverse  section  of  G  =  9 
sq.  feet,  while  the  notch  is  2  feet  wide  (i.e., 
5  =  2')  and  1  foot  deep  (A2  =  1')  (to  level  of 
surface  of  water  3  or  4  ft.  back  of  notch),  we 
have,  from  Table  C,  with  ~b  =  .60  met.  and 
A2  =  0.30  met., 

/*.  =  0.586  ; 
while  from  Table  D,  with  F  :  G  =  0.222  (or  f), 


the  ends  are 


TABLE  D. 


F 

f. 

0.05 
.10 

.000 
.000 

.15 

.001 

.20 

.003 

.25 

.007 

.30 

.014 

.35 

.026 

.40 

.044 

.45 

.070 

.50 

.107 

hence  (ft.-lb.-sec.  system  of  units),  from  eq.  (12), 


Q  =  f  x  0.586  X  1.005  X  2  X  1  X  1/64.4  X  1.0" 
=  6.30  cub.  ft.  per  second. 

CASE  IY.  Fig.  559.  Imperfect  and  incomplete  contrac- 
tion together  ;  both  end-contractions  being  "suppressed"  (by 
making  the  ends  flush  with  the  sides  of  the  reservoir,  these 
sides  being  vertical  and  1  to  the  plane  of  the  notch),  and  the 
channel  of  approach  not  being  very  deep,  i.e.,  having  a  sec- 
tional area  G  but  little  larger  than  that,  Fy  of  notch.  F=  £A2 
as  before. 

Again  we  write 

Q  =  t0A.VajX.,      .....    (13) 
with  ^.  computed  from 

......     (13)' 


*„  being  obtained  from  Table  C  ;  while 


686 


MECHANICS   OF   ENGINEERING. 


ft  =  0.041  +  0.3693    g,  ....     (13)" 

an  empirical  formula  based  by  Weisbach  on  his  own  experi- 
ments. To  save  computation,  ft  may  be  found  from  Table  E, 
founded  on  eq.  (13)". 

TABLE  E. 


F 
G~ 

.00 

.05 

.10 

.15 

.20 

.25 

.30 

.35 

.40 

.45 

.50 

0  = 

.041 

.043 

.045 

.049 

.056 

.064 

.074 

.086 

.100 

.116 

.133 

EXAMPLE.—  Fig.  559.     With 
5  =  2  ft.  (=  0.60  met.) 

ht  =  1  ft.  (—  0.30  met.), 

we  have,  from  Table  C,  >0  =  0.586. 
But,  the  ends  being  flush  with  the 
sides  of  the  reservoir  or  channel, 
and  G  being  —  6  sq.  ft.  (see  figure), 

which  is  not  excessively  large  compared  with  J^=  l/it  =  2  sq. 

ft.,  we  have,  from  Table  E,  with  F  :  G  =  0.333, 


FIG.  559. 


and  hence  [eq.  (13)  and  (13)'],  /*0  being  .586  as  in  last  example, 


Q  =  |  x  0.586  X  (1  +  .081)  X  2  X  1  X  V64.4  X  1.0 
=  6.Y8  cu.b.  ft.  per  sec. 

505.  Francis'  Formula  for  Overfalls  (i.e.,  rectangular  notches). 
— From  extensive  experiments  at  Lowell,  Mass.,  in  1851,  with 
rectangular  overfall-weirs,  Mr.  J.  B.  Francis  deduced  the  fol- 
lowing formula  for  the  volume,  Q,  of  flow  per  second  over 
such  weirs  10  feet  in  width,  and  with  A2  varying  from  0.6  to 
1.6  feet  (from  sill  of  notch  to  level  surface  of  water  a  few  feet 
back) : 


OVERFALL-WEIRS.  687 

.    .    (14) 

in  which  l>  =  width. 

This  provides  for  incomplete  contraction,  as  well  as  for  com- 
plete and  perfect  contraction,  by  making 

n  =  2  for  perfect  and  complete  contraction  (Fig.  557) ; 
n  =  1  when  one  end  only  is  flush  with  side  of  channel ; 
n  =  0  when  both  ends  are  flush  with  sides  of  channel. 

The  contraction  was  considered  complete  and  perfect  when 
the  channel  of  approach  was  made  as  wide  as  practicable, 
=  13.96  feet,  the  depth  being  about  5  feet. 

Mr.  Francis  also  experimented  with  submerged  or  "  drowned" 
weirs  in  1883  ;  such  a  weir  being  one  in  which  the  sill  is  be- 
low the  level  of  the  tail-water  (i.e.,  of  receiving  channel). 

506.  Fteley  and  Stearns' s  Experiments  at  Boston,  Mass.,  in  1877 
and  1880. — These  may  be  found  in  the  Transactions  of  the 
American  Society  of  Civil  Engineers,  vol.  xii,  and  gave  rise 
to  formulae  differing  slightly  from  those  of  Mr.  Francis  in 
some  particulars.  In  the  case  of  suppressed  end-contractions^ 
like  that  in  Fig.  559,  they  propose  formulae  as  follows : 

When  depth  of  notch  is  not  large, 

Q  (in  cub.  ft.  per  sec.)  =  3.31  IhJ  +  0.007  ~b    .    (15) 
(b  and  A2  both  infeet\ 

"A2,  the  depth  on  the  weir,  should  be  measured  from  the  sur- 
face of  the  water  above  the  curvature  of  the  sheet." 

"  Air  should  have  free  access  to  the  space  under  the  sheet." 
The  crest  must  be  horizontal.  The  formula  does  not  apply  to 
depths  on  the  weir  less  than  0.07  feet. 

When  the  depth  of  notch  is  quite  large,  a  correction  must 
be  made  for  velocity  of  approach,  <?,  thus : 

Q  (in  cub.  ft.  per  sec.)  =  3.31  &[\  + 1.5  |^-T+  0.007  b  (16) 
(b  and  A2  both  in  feet). 


688  MECHANICS    OF    ENGINEERING. 

The  channel  should  be  of  uniform  rectangular  section  for 
about  20  ft.  or  more  from  the  weir,  to  make  this  correction 
properly.  If  G  =  the  cross-section,  in  sq.  ft.,  of  the  channel 
of  approach,  c  is  found  approximately  by  dividing  an  approxi- 
mate value  of  Q  by  G  ;  and  so  on  for  closer  results. 

The  weir  may  be  of  any  length,  5,  from  5  to  19  feet. 

506a.  Recent  Experiments  on  Overfall-weirs  in  France.  —  In 

the  Annales  des  Ponts  et  Chaussees  for  October  1888  is  an 
account  of  extensive  and  careful  experiments  conducted  in 
1886  and  1887  by  M.  Bazin  on  the  flow  over  sharp-edged 
overfall-  weirs  with  end-contractions  suppressed  ;  i.e.,  like  that 
shown  in  Fig.  559.  The  widths  of  the  weirs  ranged  from 
0.50  to  2.00  metres,  and  the  depths  on  the  weirs  (A2)  from 
0.05  to  0.60  metre.  With  p  indicating  the  height  of  the  sill 
of  the  weir  from  the  bottom  of  the  channel  of  approach,  M. 
Bazin,  as  a  practical  result  of  the  experiments,  recommends 
the  following  formula  as  giving  a  reasonably  accurate  value 
for  the  volume  of  discharge  per  unit  of  time  : 


.  (IT) 

where  the  coefficient  //  has  a  value 


Eq.  (17)  is  homogeneous,  i.e.,  admits  of  any  system  of  units. 

Provision  was  made  in  these  experiments  for  the  free  en- 
trance of  air  under  the  sheet  (a  point  of  great  importance), 
while  the  walls  of  the  channel  of  approach  were  continued 
down-stream,  beyond  the  plane  of  the  weir,  to  prevent  any 
lateral  expansion  of  the  sheet.  The  value  of  p  ranged  from 
0.20  to  2.00  metres. 

Herr  Bitter  von  Wex  in  his  "  Hydrodynamik  "  (Leipsic, 
1888)  derives  formulae  for  weirs,  in  the  establishing  of  which 
some  rather  peculiar  views  in  the  Mechanics  of  Fluids  are 
advanced. 


OVERFALLS. 


Formulae  and  tables  for  discharge  through  orifices  or  over 
weirs  of  some  forms  not  given  here  may  be  found  in  the 
works  of  Weisbach,  Rankine,  and  Trautwine. 

Mr.  Hamilton  Smith,  a  rioted  American  hydraulic  engineer, 
has  recently  published  "  Hydraulics,"  a  valuable  compilation 
and  resume  of  the  most  trustworthy  experiments  in  all  fields 
of  hydraulics  (New  York,  1886:  John  Wiley  &  Sons). 

507.  Efflux  through  Short  Cylindrical  Tubes, — When  efflux 
takes  place  through  a  short  cylindrical  tube,  or  "  short  pipe," 
at  least  2^  times  as  long  as  wide, 
inserted  at  right  angles  in  the 
plane  side  of  a  large  reservoir, 
the  inner  corners  not  rounded 
(see  Fig.  560),  the  jet  issues 
from  the  tube  in  parallel  fila- 
ments and  with  a  sectional  area? 
Fm,  equal  to  that,  F,  of  interior 
of  tube. 

To  attain  this  result,  however, '  Fl»-  56°- 

the  tube  must  be  full  of  water  before  the  outer  end  is  un- 
stopped, and  must  not  be  oily ;  nor  must  the  head,  A,  be 
greater  than  about  40  ft.  for  efflux  into  the  air.  Since  at  m 
the  filaments  are  parallel  and  the  pressure-head  therefore  equal 
to  b  (=  34  ft.  of  water,  nearly),  =  that  of  surrounding  medium, 
=  head  due  to  one  atmosphere  in  this  instance ;  an  application 
of  Bernoulli's  Theorem  [eq.  (7),  §  492]  to  positions  m  and  n 
would  give  (precisely  as  in  §§  454  and  455) 


•.'!::* 
•ft;.'1 


v    =  veloc.  at  m  = 


as  a  theoretical  result  ;  but  experiment  shows  that  the  actual 
value  of  vm  in  this  case  is 


=  0.815 


(1) 


0.815  being  an  average  value  for  00 ,  the  coefficient  of  velocity,  for 
ordinary  purposes.     It  increases  slightly  as  the  head  decreases, 


690 


MECHANICS   OF   ENGINEERING. 


and  is  evidently  muck  less  than  the  value  0.97  for  an  orifice  in 
a  thin  plate,  §  495,  or  for  a  rounded  mouth-piece  as  in  §  496. 

But  as  the  sectional  area  of  the  stream  where  the  filaments 
are  parallel,  at  m,  where  vm  —  0.815  V2yA,  is  also  equal  to  that, 
F,  of  the  tube,  the  coefficient  of  efflux,  ju0  ,  in  the  formula 


is  equal  to  00  ;  i.e.,  there  is  no  contraction,  or  the  coefficient 
of  contraction,  C,  in  this  case  =  1.00. 

Hence,  for  the  volume  of  discharge  per  unit  of  time,  we 
have  practically 

'    ^_.    .     (2) 


The  discharge  is  therefore  about  £  greater  than  through  an 
orifice  of  the  same  diameter  in  a  thin  plate  under  the  same 
head  [compare  eq.  (3),  §  495]  ;  for  although  at  m  the  velocity 
is  less  in  the  present  case,  the  sectional  area  of  the  stream  is 
greater,  there  being  no  contraction. 

This  difference  in  velocity  is  due  principally  to  the  fact  that 
the  entrance  of  the  tube  has  square  edges,  so  that  the  stream 

contracts  (at  ra',  Fig.  561)  to  a 
section  smaller  than  that  of  the 
tube,  and  then  re-expands  to  the 
full  section,  F,  of  tube.  The 
eddying  and  accompanying  in- 
ternal friction  caused  by  this  re- 
expansion  (or  "sudden  enlarge- 
ment" of  the  stream)  is  the  prin- 
cipal resistance  which  diminishes 


FIG.  561. 


the  velocity.  It  is  noticeable,  also,  in 
this  case  that  the  jet  is  not  limpid  and 
clear,  as  from  thin  plate,  but  troubled 
and  only  translucent  (like  ground- 
glass).  The  internal  pressure  in  the 
stream  at  m'  is  found  to  be  less  than 
one  atmosphere,  i.e.  less  than  that  at  m, 
as  shown  experimentally  by  the  suck- 
ing inv  of  air  when  a  small  aperture  is  made  in  the  tube  op- 


Fm.  562. 


INCLINED    SHORT   PIPES.  691 

posite  m'  .  If  the  tube  itself  were  so  formed  internally  as  to 
tit  this  contracted  vein,  as  in  Fig.  562,  the  eddying  would  be 
diminished  and  the  velocity  at  in  increased,  and  hence  the 
volume  Q  of  efflux  increased  in  the  same  proportion.  (See 
§  509a.) 

If  the  tube  is  less  than  2J-  times  as  long  as  wide,  or  if  the 
interior  is  not  wet  ~by  the  water  (as  when  greasy),  or  if  the  head 
is  over  40  or  50  ft.  (about),  the  efflux  takes 
place  as  if  the  tube  were  not  there,  Fig.  563, 
and  we  have 


vm  =  0.97  Vfyh,  as  in  §  495. 

EXAMPLE.  —  The  discharge  through  a  short 
pipe  3  inches  in  diameter,  like  that  in  Fig.  560, 
is  30  cub.  ft.  per  minute,  under  a  head  of 
2'  6",  reservoir  large.  Required  the  coefficient  of  efflux 
//0  ,  ==  00  ,  in  this  case.  For  variety  use  the  inch-pound-mirt,- 
ute  system  of  units,  in  which  g  —  32.2  X  12  X  3600  (see  Note, 
§  51).  /*„  ,  being  an  abstract  number,  will  be  the  same  numer- 
ically in  any  system  of  units. 

From  eq.  (2), 

30  X  1728 


x  32  !2  x  32.2  X  12  X  60a  X  30 
=  0.803. 

508.  Inclined  Short  Tubes  (Cylindrical).—  Fig.  564.     If  the 
short  tube  is    inclined   at   some    angle 
a  <  90°    to   the   interior    plane   of  the 
reservoir  wall,  the  efflux  is  smaller  than 
when  the  angle  is  90°,  as  in  §  507. 

We  still  use  the  form  of  equation 


Q  =  fji 

but    from    Weisbach's    experiments    p  FIG.  564. 

should  be  taken  from  the  following  table: 


692  MECHANICS    OF   ENGINEERING. 

TABLE  F,  COEFFICIENT  OF  EFFLUX  (INCLINED  TUBE). 


For  cr  =    90° 
take  j.i  =  0  =  .815 

80° 
.799 

70° 

.782 

60° 
.764 

50° 

.747 

40° 
.731 

30° 
.719 

EXAMPLE. — With  h  —  12  ft.,  d  =  diam.  of  tube  —  4  ins., 
and  a  —  46°,  we  have  for  the  volume  discharged  per  sec.  (ft., 
lb.,  and  sec.) 

(>  =[0.731  +  -j-V  (.016)]|/|y  1/64.4  X 12  =1.79  cub.ft.per  sec. 

The  tube  must  be  at  least  3  times  as  long  as  wide,  to  be 
filled. 

509.  Conical  Diverging,  and  Converging,  Short  Tubes. — With 
conical  convergent  tubes,  as  at  A,  Fig.  565,  with  inner  edges 
not  rounded,  D'Aubuisson  and  Castel  found  by  experiment 
values  of  the  coefficient  of  velocity,  0,  and  of  that  of  efflux,  //, 
[from  which  the  coefficient  of  contraction,  C  =  /*  -5-  0,  may  be 


FIG.  565. 

computed,]  for  tubes  1.55  centimeters  wide  at  the  narrow  end, 
and  4.0  centimeters  long,  under  a  head  of  h  =  3  metres,  and 
with  different  angles  of  convergence.  By  angle  of  converg- 
ence is  meant  the  angle  between  the  sides  CE  and  DB,  Fig. 
565.  In  the  following  table  will  be  found  some  values  of  /* 
and  0  founded  on  these  experiments,  for  use  in  the  formulae 


vm  =  0  Vtyh     and      Q  = 
in  which  T^7 denotes  the  area  of  the  outlet  orifice  EB. 


CONICAL   SHOKT  PIPES.  693 

TABLE  G  (CONICAL  CONVERGING  TUBES). 


Angle  of  •  )  _  go  1Q, 
convergence  f 

8° 

10°  20' 

13°  30' 

19°  30' 

30° 

49° 

//  =   .895 

.930 

.938 

.946 

.924 

.895 

.847 

0=   .894 

.932 

.951 

.963 

.970 

.975 

.984 

Evidently  JJL  is  a  maximum  for 

With  a  conically  divergent  tube  as  at  MN,  having  the  in- 

^ternal  diameter  MO  =  .025    metre,  the   internal  diam.  NP 

"—  .032  metre,  and  the  angle  between  J/TTand  PO  =  4°  50'? 

Weisbach  found  that  in  the  equation  Q  =  ^F  V%gh  (where 

F  —  area  of    outlet   section  NP)  j*  should  be  =  0.553;  the 

great  loss  of  velocity  as  compared  with  V%gh  being  due  to  the 

eddying  in  the  re-expansion  from  the  contracted  section  at  M 

(corners  not  rounded),  as  occurs  also  in  Fig.  549.     The  jet  was 

much  troubled  and  pulsated  violently. 

"When  the  angle  of  divergence  is  too  great,  or  the  head  h  too 
large,  or  if  the  tube  is  not  wet  by  the  water,  efflux  with  the 
tube  filled  cannot  be  maintained,  the  flow  then  taking  place  as 
in  Fig.  563. 

Yenturi  and  Eytelwein  experimented  with  a  conically  di- 
vergent tube  (called  now  "  Ven- 
turis tube"),  with  rounded  en- 
trance to  conform  to  the  shape 
of  the  contracted  vein,  as  in 
Fig.  566,  having  a  diameter  of 
one  inch  at  mr  (narrowest  part), 
where  the  sectional  area  =  F' 
=  O.T854  sq.  in.,  and  of  1.80 
inches  at  m  (outlet),  where  area  =  F\  the  length  being  8  ins., 
and  the  angle  of  convergence  5°  9'. 

With  Q  =  pFV^yh  they  found  p  =  0.483. 

Hence  2J-  times  as  much  water  was  discharged  as  would  have 
flowed  out  under  the  same  head  through  an  orifice  in  thin 
plate  with  area  =  F'  =  the  smallest  section  of  the  divergent 
tube,  and  1.9  times  as  much  as  through  a  short  pipe  of  sec- 
tion =  F' .  A  similar  calculation  shows  that  the  velocity  at 
mf  must  have  been  vm>  =  1.55  V%gh,  and  hence  that  the  pres- 
sure at  in!  was  much  less  than  one  atmosphere. 


FIG.  566. 


694 


MECHANICS   OF   ENGINEERING. 


Mr.  J.  B.  Francis  also  experimented  with  Yenturi's  tube 
(see  "  Lowell  Hydraulic  Experiments").  See  also  p.  389  of 
vol.  6  of  the  Journal  of  Engineering  Societies,  for  experi- 
ments with  diverging  short  tubes  discharging  under  water. 
The  highest  coefficient  (yw)  obtained  by  Mr.  Francis  was  0.782. 

509a.  New  Forms  of  the  Venturi  Tube. — The  statement  made 
in  §  507,  in  connection  with  Fig.  562,  was  based  on  purely 
theoretic  grounds,  but  has  recently  (Dec.  1888)  been  com- 
pletely verified  by  experiments*  conducted  in  the  hydraulic 
laboratory  of  the  Civil  Engineering  Department  at  Cornell 
University.  Three  short  tubes  of  circular  section,  each  3  in. 
in  length  and  1  in.  in  internal  diameter  at  both  ends,  were  ex- 
perimented with,  under  heads  of  2  ft.  and  4  ft.  Call  them  A, 
B,  and  C.  A  was  an  ordinary  straight  tube  as  in  Fig.  561 ; 
the  longitudinal  section  of  B  was  like  that  in  Fig.  562,  the 
narrowest  diameter  being  0.80  in.  [see  §  495 ;  (0.8)2  =  0.64] ; 
while  C  was  somewhat  like  that  in  Fig.  566,  being  formed 
like  B  up  to  the  narrowest  part  (diameter  0.80  in.),  and  then 
made  conically  divergent  to  the  discharging  end.  The  results 
of  the  experiments  are  given  in  the  following  table : 


Name  of 
Tube. 

A 
A 

Head. 

Number 
of  Experi- 
ments. 

Range  of  Values  of  /«.. 

Average 
Values  of  /A. 

h  =  3  ft. 
ft  =  4  ft. 

4 
3 

From  0.804  to  0.823 
"    0.819  to  0.823 

0.814 
0.821 

B 
B 

li  =  2  ft. 
h  =  4  ft. 

5 

4 

"    0.875  to  0.886 
"    0.881  to  0.902 

0.882 
0.892 

C 
C 

h  =  2  f  t. 
k  =  4  ft. 

5 

4 

"    0.890  to  0.919 
"    «0.  902  to  0.923 

0.901 
0.914 

The  fact  that  B  discharges  more  than  A  is  very  noticeable, 
while  the  superiority  of  C  to  B,  though  evident,  is  not  nearly 
so  great  as  that  of  B  to  A,  showing  that  in  order  to  increase 
the  discharge  of  an  (originally)  straight  tube  (by  encroaching 
on  the  passage-way)  it  is  of  more  importance  to  fill  up  with 
solid  substance  the  space  around  the  contracted  vein  than  to 
make  the  transition  from  the  narrow  section  to  the  discharg- 
ing end  very  gradual. 

*  See  Journal  of  the  Franklin  Inst.,  for  April,  1869. 


"  FLUID   FRICTION." 

510.  "  Fluid  Friction." — By  experimenting  with  the  flow  of 
water  in  glass  pipes  inserted  in  the  side  of  a  tank,  Prof.  Rey- 
nolds of  England  has  found  that  the  flow  goes  on  in  parallel 
filaments  for  only  a  few  feet  from  the  entrance  of  the  tube, 
and  that  then  the  liquid  particles  begin  to  intermingle  and 
cross  each  other's  paths  in  the  most  intricate  manner.  To 
render  this  phenomenon  visible,  he  injected  a  fine  stream  of 
colored  liquid  at  the  inlet  of  the  pipe  and  observed  its  further 
motion,  and  found  that  the  greater  the  velocity  the  nearer  to 
the  inlet  was  the  point  where  the  breaking  up  of  the  parallel- 
ism of  flow  began.  The  hypothesis  of  laminated  flow  is, 
nevertheless,  the  simplest  theoretical  basis  for  establishing 
practical  formulae,  and  the  resistance  offered  by  pipes  to  the 
flow  of  liquids  in  them  will  therefore  be  attributed  to  the  fric- 
tion of  the  edges  of  the  laminse  against  the  inner  surface  of 
the  pipe. 

The  amount  of  this  resistance  (often  called  skin-friction) 
for  a  given  extent  of  rubbing  surface  is  by  experiment  found— 

1.  To  be  independent  of  the  pressure  between  the  liquid  and 
the  solid ; 

2.  To  vary  nearly  with  the  square  of  the  relative  velocity  ; 

3.  To  vary  directly  with  the  amount  of  rubbing  surface; 

4.  To  vary  directly  with  the  heaviness  [y,  §  409]   of   the 
liquid. 

Hence  for  a  given  velocity  v,  a  given  rubbing  surface  of 
area  =  S9  and  a  liquid  of  heaviness  y,  we  may  write 

v* 

Amount  of  friction  (force)  =  fSy  ^— ,       (1) 

*&> 

in  which/  is  an  abstract  number  called  the  coefficient  of  fluid 
friction,  to  be  determined  by  experiment.  For  a  given  liquid, 
given  character  (roughness)  of  surface,  and  small  range  of 
velocities  it  is  approximately  constant.  The  object  of  intro- 
ducing the  2^  is  not  only  because  —  is  a  familiar  and  useful 

y 

function  of  v,  but  that  v*  -f-  2^  is  a  height,  or  distance,  and  there- 
fore the  product  of  S  (an  area)  by  v*  -r-  2g  is  a  volume,  and  this 
volume  multiplied  by  y  gives  the  weight  of  an  ideal  prism  of 


MECHANICS   OF   ENGINEERING. 


the  liquid ;  hence  S  -  -  y  is  &  force  and  f  must  be  an  abstract 
Zg 

number  and  therefore  the  same  in  all  systems  of  units,  in  any 
given  case  or  experiment. 

In  his  experiments  at  Torquay,  England,  the  late  Mr.  Froude 
found  the  following  values  for/",  the  liquid  being  salt  water, 
while  the  rigid  surfaces  were  the  two  sides  of  a  thin  straight 
wooden  board  -f$  of  an  inch  thick  and  19  inches  high,  coated 
or  prepared  in  various  ways,  and  drawn  edgewise  through  the 
water  at  a  constant  velocity,  the  total  resistance  being  measured 
by  a  dynamometer. 

511.  Mr.  Froude' s  Results. — (Condensed.)  [The  velocity  was 
the  same  =  10  ft.  per  sec.  in  each  of  the  following  cases.  For 
other  velocities  the  resistance  was  found  to  vary  nearly  as  the 
square  of  the  velocity,  the  index  of  the  power  varying  from 
1.8  to  2.16.] 

TABLE  H. 


Character  of  Surface. 

Value  of  /  [from  eq.  (1),  §  510]. 

Varnish  /  — 

2  ft.  long. 

When  the 
8  ft.  long. 

board  was 
20  ft.  long. 

50  ft.  long. 

0.0041 
.0038 
.0030 
.0087 
.0081 
.0090 
.0110 

0.0032 
.0031 
0028 
.0063 
.0058 
.0062 
.0071 

0.0028 
.0027 
.0026 
.0053 
.0048 
.0053 
.0059 

0.0025 

Paraffin  e     . 

Tinfoil  

.0025 
.0047 
.0040 
.0049 

Calico  

Fine  Sand 

Medium  Sand  

Coarse  Sand  

N.B.  These  numbers  multiplied  by  100  also  give  the  mean  f  fictional  resistance  in 
Ibs.  per  sq.  foot  of  area  of  surface  in  each  case  (v  =  10'  per  sq.  sec.),  considering  the 
heaviness  of  sea  water,  64  Ibs.  per  cubic  foot,  to  cancel  the  2g  =  64.4  ft.  per  sq.  sec.  of 
eq.  (1)  of  the  preceding  paragraph. 

For  use  in  formulae  bearing  on  flow  in  pipes,  f  is  best  deter- 
mined directly  by  experiments  of  that  very  nature,  the  results 
of  which  will  be  given  as  soon  as  the  proper  formulae  have  been 
established. 

512.  Bernoulli's  Theorem  for  Steady  Flow,  with  Friction.— [The 
student  will  now  re-read  the  first  part  of  §  492,  as  far  as  eq. 
(1).]  Considering  free  any  lamina  of  fluid,  Fig.  567,  (according 
to  the  subdivision  of  the  stream  agreed  upon  in  §  492  referred 


BERNOULLI'S  THEOREM  WITH  FRICTION. 


697 


to,)  the  frictions  on  the  edges  are  the  only  additional  forces  as 
compared  with  the  system  in  Fig. 
534.  Let  w  denote  the  length 
of  the  wetted  perimeter  of  the 
base  of  this  lamina  (in  case  of  a 
pipe  running  full,  as  we  here 
postulate,  the  wetted  perimeter 
is  of  course  the  whole  perimeter^ 
but  in  the  case  of  an  open  chan" 
nel  or  canal,  w  is  only  a  portion 
of  the  whole  perimeter  of  the 
cross-section).  Then,  since  the 
area  of  rubbing  surface  at  the  edge  is  S—  wdsf,  the  total  fric- 
tion for  the  lamina  is  [by  eq.  (1),  §  510]  =fwy  (v*  -r-  %g)ds' . 
Hence  from  vdv  =  (tan.  accel.)  X  ds,  and  from  (tan.  accel.)  — 
[^"(tang.  compons.  of  acting  forces)]  -f-  (mass  of  lamina),  we 
have 

v* 
Fp  —  F(p  -\-  dp)  -f-  Fyds'  cos  0  —fwy  —  dsf 

vdv  = ^ — j-. .ds...(a) 

Fyds'  -^  g 

As. in  §  492,  so  here,  considering  the  simultaneous  advance  of 
all  the  laminae  lying  between  any  two  sections  m  and  n  during 
the  small  time  dt,  putting  ds'  =  ds,  and  dsf  cos  0  =  —  dz  (see 
Fig.  568),  we  have,  for  any  one  lamina, 


—  vdv,  -\ —  dp -\-dz-  — 
g  Y 


(1)'' 


fi1  *  9, 

-/.  ^J 

Now  conceive  an  infinite  number  of  equations  to  be  formed 
like  eq.  (1),  one  for  each  la- 
mina between  n  and  m,  for  the 
same  dt,  viz.,  a  dt  of  such 
length  that  each  lamina  at  the 
end  of  dt  will  occupy  the 
same  position,  and  acquire  the 
same  values  of  v,  z,  and  p, 
that  the  lamina  next  in  front 
had  at  the  beginning  of  the  FIG.  568. 

dt  (this  is  the  characteristic  of  a  steady  flow).     Adding  up 


698  MECHANICS   OF   ENGINEERING. 

the  corresponding  terms  of  all  these  equations,  we  have  (re- 
membering that  for  a  liquid  y  is  the  same  in  all  laminae), 

1     f*m  1    f*m  (*m  -f        f*m  <?/?  / 

1  /  vdv  +  -  /  dp  +  /  dz  =  -  i-  .   /    ~v\ls ;   .  (2)  '' 

qJn  yt/n    •*      '     Jn  ^y       Un      J-T 

i.e.,  after  transposition  and  writing  It  for  F-±-  w,  for  brevity, 


This  is  Bernoulli  's  Theorem  for  steady  flow  of  a  liquid  in 
a  pipe  of  slightly  varying  sectional  area  F,  and  internal  perim- 
eter w,  taking  into  account  no  resistances  or  friction,  except 
the  "  skin-friction,"  or  u  fluid-friction,"  of  the  liquid  and  sides 
of  the  pipe. 

Resistances  due  to  the  internal  friction  of  eddying  occasioned 
by  sudden  enlargements  of  the  cross-section  of  the  pipe,  by 
elbows,  sharp  curves,  valve-gates,  etc.,  will  be  mentioned  later. 
The  negative  term  on  the  right  in  (3)  is  of  course  a  height  or 
head  (one  dimension  of  length),  as  all  the  other  terms  are  such, 
and  since  it  is  the  amount  by  which  the  sum  of  the  three  heads 
(viz.,  velocity-head,  pressure-head,  and  potential  head)  at  m, 
the  down-stream  position,  lacks  of  being  equal  to  the  sum  of 
the  corresponding  heads  at  n^  the  up-stream  position  or  section, 
we  may  call  it  the  "Loss  of  Head"  due  to  skin-friction  between 
n  and  m\  also  called  friction-head,  or  resistance-head,  or 
height  of  resistance. 

The  quantity  R  =  F  ~-  w  =  sectional-area  -^  wetted-pe 
rimeter,  is  an  imaginary  distance  or  length  called  the  Hydrau- 
lic Mean  Radius,  or  Hydraulic  Mean  Depth,  or  simply 
hydraulic  radius  of  the  section.  For  a  circular  pipe  of  diam- 
eter =  J 


while  for  a  pipe  of  rectangular  section, 
r>_       rib 


FRICTION  IN   PIPES. 


699 


-':.-.'.••.  :  :•.:? 

—    —  —  ~ 

•i 

—                           ^ 

r                                   i 

z^r-^^ 

i 

TI  T!H 

^'. 

—  "- 

y 

"•>?•:  -\ 

=>^ 

_ 

^  ^  3 

r  -j}.  --_•  —  •  •  —  „  #  :  * 

=-==( 

"~  .?& 

=r~  r  

FIG.  569. 

(513.  Problems    involving  Friction-heads;    and   Examples  of 
Bernoulli's  Theorem  with  Friction. 

PROBLEM  I. — Let  the  portion  of  pipe  between  n  and  m  be 
level,  and  of  uniform  cir-        ^..~... 
cular  section  and  diameter 
=  d.    The  jet  at  m  dis- 
charges into   the  air,  and 
has  the  same  sectional  area, 
F=  ^-7r^2,as  the  pipe;  then 
the  pressure-head  at  m  is 

34    feet    (for 


^  =  £  = 
r 

water),  and  the  velocity- 
head  at  m  is  =  that  at  n,  since  vm  =  vn .     The  height  of  the 
water  column  in  the  open  piezometer  at  n  is  noted,  and  =  yn 

(so  that  the  pressure-head  at  n  is  S^L  =  yn  +  b) ;    while   the 

r 

length  of  pipe  from  n  to  m  is  —  I. 

Knowing  I,  d,  yn,  and  having  measured  the  volume  Q,  of 
flow,  per  unit  of  time,  it  is  required  to  find  the  form  of  the 
friction-head  and  the  value  off.  From 

Fmvm  =  Q,    or    %7fd?vm  =  Q,  '.     .    .     .    (1) 

vm  becomes  known.  Also,  vm  is  known  to  be  =  vn,  and  the 
velocity  at  each  ds  is  v  ~  vm ,  since  F  (sectional  area)  is  con- 
stant along  the  pipe,  and  Fv  —  Q.  The  hydraulic  radius  is 


(2) 


the  same  for  all  the  ds's  between  n  and  m. 

Substituting  in  eq.  (3)  of  §  512,  with  the  horizontal  axis  of 
the  pipe  as  a  datum  for  potential  heads,  we  have 

2  o  /»  2       /•m 

—  4- 5  -I-  0  =  — ^-  4-  yn -4- b -4- 0  —  £==  .  —  I  ds:   .    (3) 
zg  %g  ^d     2</  ^n 

Pm 

i.e.,  since  J^  ds  =  I  =  length  of  pipe  from  n  to  m,  the  friction- 
head  for  a  pipe  of  length  =  I,  and  uniform  circular  section 
qf  diameter  =  d,  reduces  to  the  form 


AV| 


700 


MECHANICS   OF   ENGINEERING. 


Friction-head  =  4:f-^  .  — -;     ....     (4) 
c  d     ty 

where  v  =  velocity  of  water  in  the  pipe,  being  in  this  case 
also  =  vm  and  =  vn .  Hence  this  friction-head  varies  directly 
as  the  length  and  as  the  square  of  the  velocity,  and  inversely 
as  the  diameter  •  also  directly  as  the  coefficient/. 

From  (3),  then,  we  derive  (for  this  particular  problem) 

Piezometer-height  at  n  —  yn  =  4/—  .  — ;  .     .     (5) 

i.e.,  the  open  piezometer-height  at  n  is  equal  to  the  loss  of  head 
(all  of  which  is  friction-head  here)  sustained  between  n  and  the 
mouth  of  the  pipe.  (Pipe  horizontal.) 

EXAMPLE. — Required  the  value  of/*,  knowing  that  d  —  3  in., 
yn  (by  observation)  =  10.4  ft.,  and  Q  =  0.1960  cub.  ft.  per 
sec.,  while  I  =  400  ft.  (n  to  m).  From  eq.  (1)  we  find,  in  ft.- 
Ib.-sec.  system,  the  velocity  in  the  pipe  to  be 

4$       4  X  0.1960 

v  =  ^  =  -%r      -  =  4.0  ft.  per  sec.; 


then,  using  eq.  (5),  we  determine  f  to  be 


-  tgynd  _  2  X  32.2  X  \  X  10.4 

-  ~4hT      Tx  wo  x  4~ 


PROBLEM  II.  Hydraulic  Accumulator.  —  Fig.  570.     Let  the 
area  Fn  of  the  piston  on  the  left  be  quite  large  compared  with 


ffif 


FIG.  570. 


that  of  the  pipes  and  nozzle.     The  cylinder  contains  a  friction- 


FRICTION-HEAD   IN   PIPES.  701 

less  weighted  piston,  producing  (so  long  as  its  downward  slow 
motion  is  uniform)  a  fluid  pressure  on  its  lower  face  of  an 
intensity  pn  =  [G-  -{-  Fnpa\  -f-  Fn  per  unit  area  (pa  —  one 
atmos.). 

Hence  the  pressure-head  at  n  is 


(6) 


where  G  =  load  on  piston. 

The  jet  has  a  section  at  m  =  Fm  =  that  of  the  small  straight 
nozzle  (no  contraction).  The  junctions  of  the  pipes  with  each 
other,  and  with  the  cylinder  and  nozzle,  are  all  smoothly 
rounded  ;  hence  the  only  losses  of  head  in  steady  flow  between 
n  and  m  are  the  friction-heads  in  the  two  long  pipes,  neglect- 
ing that  in  the  short  nozzle.  These  friction-heads  will  be  of 
the  form  in  eq.  (4),  and  will  involve  the  velocities  vt  and  v^ 
respectively  in  these  pipes'  (supposed  running  full).  iol  and  v9 
may  be  unknown  at  the  outset,  as  here. 

Knowing  G  and  all  dimensions  and  heights,  we  are  required 
to  find  the  velocity  vm  of  the  jet,  flowing  into  the  air,  and  the 
volume  of  flow,  Q,  per  unit  of  time,  assuming  f  to  be  known 
and  to  be  the  same  in  both  pipes  (not  strictly  true). 

Let  the  lengths  and  diameters  be  denoted  as  in  Fig.  570, 
their  sectional  areas  F^  and  F9  ,  the  unknown  velocities  in  them 
vt  and  vt  . 

From  the  equation  of  continuity  [eq.  (3),  §  490],  we  have 


and     «.=  .       ...     (7) 

"** 


To  find  vmi  we  apply  Bernoulli's  Theorem  (with  friction), 
eq.  (3),  §  512,  taking  the  down-stream  position  m  in  the  jet 
close  to  the  nozzle,  and  the  up-stream  position  n  just  under  the 
piston  in  the  cylinder  where  the  velocity  vn  is  practically  noth- 
ing. Then  with  m  as  datum  plane  we  have 


=  0  +       +  A-V.-V..    (8) 


702  MECHANICS    OF   ENGINEERING. 

Apparently  (8)  contains  three  unknown  quantities,  vm ,  vl , 
and  v3 ;  but  from  eqs.  (7)  v,  and  v9  can  be  expressed  in  terms 
of  vm ,  whence  [see  also  eq.  (6)] 


or,  finally, 

;  -  .  (io) 


and  hence  we  have  also 

Q  =  FA (11) 

EXAMPLE. — If  we  replace  the  force  G  of  this  problem  by 
the  thrust  P  exerted  along  the  pump-piston  of  a  steam  fire- 
engine,  we  may  treat  the  foregoing  as  a  close  approximation 
to  the  practical  problem  of  such  an  apparatus,  the  pipes  being 
consecutive  straight  lengths  of  hose,  in  which  (for  the  probable 
values  of  vl  and  v^)  we  may  takey  =  .0075  (see  '^ire-streams," 
by  Geo.  Ellis,  Springfield,  Mass.).  (Strictly,  f  varies  somewhat 
with  the  velocity;  see  §  517.)  Let  P=  12000  Ibs.,  and  the 
piston-area  at  n  =  Fn  —  72  sq.  in.  =  J  sq.  ft.  Also,  let  h  =  20 
ft.,  and  the  dimensions  of  the  hose  be  as  follows : 

dt  =  3  in.,     d^  =  2  in.,     dn  (of  nozzle)  =  1  in. ; 
^  =  400  ft.,     lt  =  500  ft. 

"With  the  foot-pound-second  system  of  units,  we  now  have 
[eq.  (10)] 


+  4X 


FRICTION-HEAD    IN   PIPES.  703: 


__      /2  X  32.2  X  404  . 


0.59  +  5.62 

i.e.,  vm  =  60.0  ft.  per  sec.     If  this  jet  were  directed  vertically 

v  2 
upward  it  should  theoretically  attain  a  height  —  -^-  =  nearly 

56  feet,  but  the  resistance  of  the  air  would  reduce  this  to  about. 
40  or  45  ft. 

We  have  further,  from  eq.  (1), 


Q  =  Fmvm  =  x  60.0  =  3.27  cub.  ft.  per,sec. 


If  there  were  no  resistance  in  the  hose  we  should  have,  from; 
§497, 


vm  =*  / 

y 


+  A   =  V2  X  32.2  x  404  =  161.3  ft.  per  sec.. 


513a.  Influence  of  Changes  of  Temperature.  —  Although  Poi- 
seuille  and  Hagen  found  that  with  glass  tubes  of  very  small 
diameter  the  flow  of  water  was  increased  threefold  by  a  rise  of 
temperature  from  0°  to  45°  Cent.,  it  is  unlikely  that  with  com- 
mon pipes  the  rate  of  flow  is  appreciably  affected  by  the  ordi- 
nary fluctuations  of  temperature;  at  any  rate,  experiments  of 
sufficient  precision  are  wanting,  as  regards  such  an  influence. 
See  Mr.  Hamilton  Smith's  "  Hydraulics,"  p.  16,  where  lie 
says  :  u  Changes  by  variation  in  T  (temperature)  will  probably 
only  be  appreciable  with  small  orifices,  or  with  very  low  heads. 
for  orifices  or  weirs." 

514.  Loss  of  Head  in  Orifices  and  Short  Pipes.  —  So  long  as  the 
steady  flow  between  two  localities  n  and  m  takes  place  in  a  pipe 
having  no  abrupt  enlargement  or  diminution  of  section,  nor 
sharp  curves,  bends,  or  elbows,  the  loss  of  head  may  be  ascribed 
solely  to  the  surface  action  (or  "  skin-friction")  between  water 
and  pipe  ;  but  the  introduction  of  any  of  the  above-mentioned 
features  occasions  eddying  and  internal  disturbance,  and  fric- 
tion (and  consequent  heat)  ;  thereby  causing  further  deviations 


704  MECHANICS   OF   ENGINEERING. 

from  Bernoulli's  Theorem ;  i.e.,  additional  losses  of  head,  or 
heights  of  resistance. 

From  the  analogy  of  the  form  of  a  friction-head  in  a  long 
pipe  [eq.  (4),  §  513],  we  may  assume  that  any  of  the  above 
heights  of  resistance  is  proportional  to  the  square  of  the  veloc- 
ity, and  may  therefore  always  be  written  in  the  form 

j  Loss  of  Head  due  to  any  \ «.  v*  ^ 

(  cause  except  skin-friction  f  ~       2#~J *  ' 

in  which  v  is  the  velocity  of  the  water  in  the  pipe  at  the  sec- 
tion where  the  resistance  occurs ;  or  if,  on  account  of  an 
abrupt  enlargement  of  the  stream-section,  there  is  a  correspond- 
ing diminution  of  velocity,  then  v  is  always-  to  denote  this 
diminished  velocity  (i.e.,  in  the  down-stream  section).  This 
velocity  v  is  often  an  unknown  at  the  outset. 

C,  corresponding  to  the  abstract  factor  4/  — -  in  the  height  of 

ct 

resistance  due  to  skin-friction  [eq.  (4),  §  513],  is  an  abstract 
number  called  the  Coefficient  of  Resistance,  to  be  determined 
experimentally ;  or  computed  theoretically,  where  possible. 
Roughly  speaking,  it  is- independent  of  the  velocity,  for  a  given 
fitting,  casing,  pipe-joint,  elbow,  bend,  valve-gate  at  a  definite 
opening,  etc.,  etc. 

515.  Heights  of  Resistance  (or  Losses  of  Head)  Occasioned  by 
Short  Cylindrical  Tubes. — When  dealing  with  short  tubes  dis- 
charging into  the  air,  in  §  507,  deviations  from  Bernoulli's 
•Theorem  were  made  good  by  using  a  coefficient  of  velocity  0, 
dependent  on  experiment.  This  device  answered  every  pur- 
pose for  the  simple  circumstances  of  the  case,  as  well  as  for 
simple  orifices.  But  the  great  variety  of  possible  designs  of  a 
compound  pipe  (with  skin-friction,  bends,  sudden  changes  of 
cross-section,  etc.)  renders  it  almost  impossible,  in  such  a  pipe, 
to  provide  for  deviations  from  Bernoulli's  Theorem  by  a  single 
coeificient  of  velocity  (velocity  of  jet,  that  is)  for  the  pipe  as  a 
whole,  since  new  experiments  would  be  needed  for  each  new 
design  of  pipe.  Hence  the  great  utility  of  the  conception  of 
"  loss  of  head,"  one  for  each  source  of  resistance. 


LOSS   OF  HEAD   IN   A   SHOUT  PIPE.  705 

If  a  long  pipe  issues  from  the  plane  side  of  a  reservoir  and 
the  corners  of  the  junction  are  not  rounded  [see  Fig.  571],  we 

shall  need  an  expression   for 

— --,  — •*'    the  loss  of  head  at   the   en- 

a  ft    trance,  E,  as  well  as  that 


'2 

Fia  571'  due  to  the  skin-friction  in  the 

pipe.  But,  whatever  the  velocity,  v,  in  the  pipe  is  going  to 
be,  influenced  as  it  is  both  by  the  entrance  loss  of  head  and 
the  skin-friction  head  (in  applying  Bernoulli's  Theorem),  the 

/y» 

loss  of  head  at  E,  viz.,  £E  —  ,  will  be  just  the  same  as  if  efflux 

2g 

took  place  through  enough  of  the  pipe  at  E  to  constitute  a 
"  short  pipe,"  discharging  into  the  air,  under  some  head  h 
(different  from  h'  of  Fig.  571)  sufficient  to  produce  the  same 
velocity  v.  But  in  that  case  we  should  have 


(1) 


(See  §§  507  and  508,  0  being  the  "  coefficient  of  velocity,"  and 
h  the  head,  in  the  cases  mentioned  in  those  articles.) 

We  therefore  apply  Bernoulli's  Theorem  to  the  cases  of 
those  articles  (see  Figs.  560  and  564:)  in  order  to  determine  the 
loss  of  head  due  to  the  short  pipe,  and  obtain  (with  m  as  datum 
level  for  potential  heads) 


(2) 


Now  the  v  of  eq.  (2)  is  equal  to  the  vm  of  the  figures  referred 
to?  and  £#  is  a  coefficient  of  resistance  for  the  short  pipe,  and 
we  now  desire  its  value.  Substituting  for 


706  MECHANICS   OF   ENGINEERING. 

its  value  02A  from  eq.  (1),  we  have 

1 


(3) 


Hence  when  a  =  90°  (i.e.,  the  pipe  is  ~]  to  the  inner  reser- 
voir surface),  we  derive 


and  similarly,  for  other  values  of  a  (taking  0  from  the  table, 
§  508),  we  compute  the  following  values  of  £E  (corners  not 

rounded)  for  use  in  the  expression  for  "  loss  of  head,"  £E  — : 


For  a  =   90° 

80° 
.565 

70° 
.635 

60° 
.713 

50° 
.794 

40° 
.870 

30° 

.987 

From  eq.  (4)  we  see  that  the  loss  of  head  at  the  entrance  of 
the  pipe,  corners  not  rounded,  with  a  =.  90°,  is  about  one  half 
(.505)  of  the  height  due  to  the  velocity  v  in  that  part  of  the 
pipe  (v  being  the  same  all  along  the  pipe  if  cylindrical).  The 
value  of  v  itself,  Fig.  571,  depends  on  all  the  features  of  the 
design  from  reservoir  to. nozzle.  See  §  518. 

If  the  corners  at  E  are  properly  rounded,  the  entrance  loss  of 
head  may  practically  be  done  away  with;  still,  if  v  is  quite 
small  (as  it  may  frequently  be,  from  large  losses  of  head 
farther  down-stream),  the  saving  thus  secured,  while  helping 
to  increase  v  slightly  (and  thus  the  saving  itself),  is  insignifi 
cant. 

516.  General  Form  of  Bernoulli's  Theorem,  considering  all 
Losses  of  Head, 

In  view  of  preceding  explanations  and  assumptions,  we  may 
write  in  a  general  and  final  form  Bernoulli's  Theorem  for  a 
steady  flow  from  an  up-stream  position  n  to  a  down-stream 
position  m,  as  follows : 

a  a  (  all  losses  of  head  ) 

?±-+S»  +  gm  =  ?!L  +  £*  +  en-  \occurringletween\ 
ty        Y  fy        r  I          nandm 


LOSSES   OF  HEAD    IN   GENERAL.  707 

Each  loss  of  head  (or  height  of  resistance)  will  be  of  the  form 

v*  I   v* 

£  — .  (except  skin-friction  head  in  long  pipes,  viz.,  4/  — -  — 

£i(J  (JL    +-t(J 

the  v  in  each  case  being  the  velocity,  known  or  unknown,  in 
that  part  of  the  pipe  where  the  resistance  occurs  (and  hence 
is  not  necessarily  equal  to  vm  or  vn). 

517.  The  Coefficient,/,  for  Friction  of  Water  in  Pipes.— See 
eq.  (1),  §  510.  Experiments  have  been  made  by  Weisbach, 
Eytelwein,  Darcy,  Bossut,  Prony,  Dubuat,  Fanning,  and  oth- 
ers, to  determine  /  in  cylindrical  pipes  of  various  materials 
(tin,  glass,  zinc,  lead,  brass,  cast  and  wrought  iron)  of  diameters 
from  J  inch  up  to  36  inches.  In  general,  the  following  deduc- 
tions may  be  made  from  these  experiments : 

1st.  /  decreases  when  the  velocity  increases ;  e.g.,  in  one 
case  with  the 

same  pipe/  was  —  .0070  for  v  =  2'  per  sec., 
while/  was  =  .0056  for  v  =  20'  per  sec. 

2dly.  /  decreases  slightly  as  the  diameter  increases  (other 
things  being  equal); 

e.g.,  in  one  experiment/ was  =  .0069  in  a  3-in.  pipe, 
while  for  the  same  velocity/ was  —  .0064  in  a  6-in.  pipe. 

3dly.  The  condition  of  the  interior  surface  of  the  pipe 
affects  the  value  of/,  which  is  larger  with  increased  roughness 
of  pipe. 

Thus,  Darcy  found,  with  afoul  iron  pipe  with  d  =  10  in. 
and  veloc.  =  3.67  ft.  per  sec.,  the  value  .0113  for/;  whereas 
Fanning  (see  p.  238  of  his  "  Water-supply  Engineering"),  with 
a  cement-lined  pipe  and  velocity  of  3.74  ft.  per  sec.  and  d  = 
20  inches,  obtained/  =  .0052. 

Weisbach,  finding  the  first  relation  very  prominent,  pro- 
posed the  formula 

.00429 

/=  0.00359  +      ,    ..    —  ===== 
Vv  (in  ft.  per  sec.) 

when  the  velocities  are  great,  while  Darcy,  taking  into  account 
both  the  1st  and  2d  relations  above,  writes  (see  p.  585,  Ran- 
kine's  Applied  Mechanics) 


708  MECHANICS   OF   ENGINEERING. 

/=  .0043  [~1  -I  --  *  ---  1 
L  ~  '-9  X  di&in.  in  ft  J 


ft.  per  sec.  _         is  x  diam.  in 


i 

ft.J 


For  practical  purposes,  Mr.  J.  T.  Fanning  has  recommended, 
and  arranged  in  an  extensive  table  (pp.  242-246  of  his  book 
just  mentioned),  values  of  f  for  clean  iron  pipe,  of  diameters 
from  J  inch  to  96  inches,  and  for  velocities  of  0.1  ft.  to  20  ft. 
per  second.  Of  this  the  table  opposite  is  an  abridgment,  in- 
serted with  Mr.  Fanning's  permission,  for  use  in  solving  nu- 
merical problems. 

In  obtaining/  for  slightly  tuberculated  and  for  foul  pipes, 
the  recommendations  of  Mr.  Fanning  seem  to  justify  the  fol- 
lowing rules  : 


For  slightly  tuberculated  pipes  of  diams.  =  j-  ft. 


and  for  foul  pipes  of  same  size 


we  should  add  23$ 


1ft. 

34$ 
60$ 


2ft. 

16$ 
38$ 


4ft. 

13$ 
25$ 


of  the/*  for  clean  pipes,  to  itself.  For  example,  if/  =  .007 
for  a  certain  ^-ft.  pipe  when  clean,  with  velocity  =  0.64  ft.  per 
sec.,  we  have/=  .007  X  1.72  =  .01204  when  it  is  foul. 

For  first  approximations  a  mean  value  of  f  =  .006  may  be 
employed,  since  in  some  problems  sufficient  data  may  not  be 
known  in  advance  to  enable  us  to  take/ from  the  table. 

EXAMPLE.- — Fig.  572.  In  the  steady  pumping  of  crude 
petroleum  weighing  y  =  55  Ibs.  per  cubic  foot,  through  a  six- 
inch  pipe  30  miles  long, 
to  a  station  700  ft.  higher 
than  the  pump,  it  is 
found  that  the  pressure 
in  the  pump  cylinder  at 
n,  necessary  to  keep  up 
a  velocity  of  4.4  ft.  per 
sec.  in  the  pipe,  is  1000 

Ibs.  per  sq.  inch.  Kequired  the  coefficient  f  in  the  pipe.  As 
all  losses  except  the  friction-head  in  the  pipe  are  insignificant, 
the  latter  only  will  be  considered.  The  velocity-head  at  n  may 


TABLE   OF  VALUES   OF/. 


709 


« 

I 


O      .3 

1 


a  -£ 

I } 

s    ^ 

H 
fc 

w 

3 

I 

§ 

o 
H 


s.s< 


sag 


S  S 


111 


Ba'sa 


II 

II  II 


OS      CO 

$  % 

8 


T-l        CO        O        J>        O        T-l 


OS     GO 

g  3 


OS     GO     Ci     O     i> 

O     OS     GO     t-     IO 
10     ^     TH     T*     ^ 


1     £ 


10     10     10     Ttf     -rtf 


CO         T-l 

OS     GO 
10      10 


O     CO     ^     O     O     TH     IO 


•^     CO     T— I      O     OS 
IO     IO     IO     IO     -^ 


^     CO     b-     OS 

O     OS     t-     JO 

'.—     CO     CO     CO 


8 


(M     -*     C5     CO 

CO      ^       CQ      T-H 

IO     1O     1O     IO 


§  %  8 

IO     IO     IO 


O     T-I     IO     *>•     O     C5     OS 

IO  CO  O  GO  CO  IO  "tfl 
CO  CO  CO  IO  O  IO  IO 


^     t-     CO 
GO     CO     Tt< 

t-    t-    t- 


/   IO 
CO 


§^ 


8 


TH  O  GO  O  -—I 
C~  -*  T-I  OS  GO 
CO  CO  CO  IO  O 


OS  CO  ^ 

TH  TH  O 

TH  T-l  T-l 

o  o  o 


J>     CO     CO     CO 


o    t>   -*    o 

»O     CO     <M     T-I 

TH        T-l         TH         TH 

0000 


O     00     IO 

CO      TH      T— I 
CO     <£>     CO 


TH     CO     CO     O 
O     O     O     TH 


0  IO 

01  « 


o    o    o    o    o    o    o 

00     ^     CO     GO     C<1     CO     O 


710  MECHANICS   OF  ENGINEERING. 

be  put  =  0  ;  the  jet  at  m  being  of  the  same  size  as  the  piper 
the  velocity  in  the  pipe  is  =vm,  and  therefore  vm  =  4.4  ft.  per 
sec.  Notice  that  m,  the  down-stream  section,  is  at  a  higher 
level  than  n. 

From  Bernoulli's  Theorem,  §  516,  we  have,  with  n  as  a 
datum  level, 


Using  the  ft.,  lb.,  and  sec.,  we  have 

h  =  700  ft.,    vm2  +  2g  =  0.30  ft., 
while 

1000 


and    .  = 


_ 

55  y  55 

Hence,  in  eq.  (1), 

30  X  5280    (4.4) 


-0.30  +  38.5  +  700  =  2618  -  4/. 


64.4 


Solving  for/*,  we  have  f  =  .00485  (whereas  for  water,  with 
v  =  4A  ft.  per  sec.  and  d=%  ft.,  the  table,  p.  146,  gives 
/=.  00601. 

If  the  y  of  the  petroleum  had  been  50  Ibs.  per  cubic  foot, 

instead  of  55,  we  would  have  obtained  £?  =  2880  feet  and  f 

Y 
=  .0056. 

518.  Flow  through  a  Long  Straight  Cylindrical  Pipe,  including 
both  friction-head  and  entrance  loss  of  head  (corners  not  rounded); 

reservoir  large.    Fig.  573. 
The  jet    issues    directly 
I    from  the  end  of  the  pipe, 
!    in  parallel  filaments,  into 

*•  |  A  7 

therefore 

.-"..v  Ijas  same  section  as  pipe; 
hence,  also,  vm  of  the  jet 


E 

FIG.  573. 

=  v  in  the  pipe  (which  is  assumed  to  be  running  full),  and  is 


COEFFICIENT   OF   LIQUID   FRICTION.  711 


therefore  the  velocity  to  be  used  in  the  loss  of  head  C#  —  -  at 

the  entrance  ^(§  515). 

Taking  m  and  n  as    in    figure   and   applying   Bernoulli's 
Theorem  (§  474),  with  m  as  datum  level  for  the  potential  heads 


zm  and  zn  ,  we  have 


-<«  .    (1) 


Three  different  problems  may  now  be  solved  : 

First,  required  the  head  li  to  keep  up  a  flow  of  given  volume 

=  Q  per  unit  of  time  in  a  pipe  of  given  length  I  and  diameter 

=  d. 
From  the  equation  of  continuity  we  have 

Q  =  Fmvm  =  %nd*vm  ; 

4:0 

.'.  veloc.  of  jet,  which  =  veloc.  in  pipe,  =  vm  =  —~.  .     .     (2) 


Having  found  vm  =  v,  from  (2),  we  obtain  from  (1)  the  re- 
quired A,  thus  : 


Now  G,  =  0.505  if  a  =  90°  (see  §  515),  while  /  may  be 
taken  from  the  table,  §  517,  for  the  given  diameter  and  com- 
puted velocity  [vm  =  i>,  found  in  (2)],  if  the  pipe  is  clean  ;  if 
not  clean,  see  end  of  §  517,  for  slightly  tuberculated  and  for 
foul  pipes. 

Secondly.  Given  the  head  A,  and  the  length  I  and  diameter 
d  of  pipe,  required  the  velocity  in  the  pipe,  viz.,  v,  =  vm  ,  that 
of  jet  ;  also  the  volume  delivered  per  unit  of  time,  Q.  Solv- 
ing eq.  (1)  for  vm,  we  have 


d 


712  MECHANICS   OF   ENGINEERING. 

whence  Q  becomes  known,  since 

(5) 


[NOTE.  —  The  first  radical  in  (4)  might  for  brevity  be  called 
a  coefficient  of  velocity,  0,  for  this  case.  Since  the  jet  has  the 
same  diameter  as  the  pipe,  this  radical  may  also  be  called  a 
coefficient  of  efflux.'} 

Since  in  (4)  f  depends  on  the  unknown  v  as  well  as  on  the 
known  d,  we  must  first  put/"  —  .006  for  a  first  approximation 
for  vm  ;  then  take  a  corresponding  value  for  f  and  substitute 
again  ;  and  so  on. 

Thirdly,  knowing  the  length  of  pipe  and  the  head  h,  we 
wish  to  find  the  proper  diameter  d  for  the  pipe  to  deliver  a 
given  volume  Q  of  water  per  unit  of  time.  Now 


which  substituted  in  (1)  gives 


that  is, 


As  the  radical  contains  d,  we  first  assume  a  value  for  d, 
with/*—  .006,  and  substitute  in  (7).  With  the  approximate 
value  of  d  thus  obtained,  we  substitute  again  with  a  new  value 
for  f  based  on  an  approximate  v  from  eq.  (6)  (with  d  —  its 
first  approximation),  and  thus  a  still  closer  value  for  d  is  de- 
rived ;  and  so  on.  (Trautwine's  Pocket-book  contains  a  table 
of  fifth  roots  and  powers.)  If  I  is  quite  large,  we  may  put 
d  =  0  for  a  first  approximation.  In  connection  with  these 
examples,  see  last  figure. 


LONG   PIPES.  713 

EXAMPLE  1.  —  What  head  h  is  necessary  to  deliver  120  cub. 
ft.  of  water  per  minute  through  a  clean  straight  iron  pipe  140 
ft.  long  and  6  in.  in  diameter  ? 

From  eq.  (2),  with  ft.,  lb.,  and  sec.,   we  have 

v  =  vm  =  [4  X  W]  *  *(*)'=  10-18  ft-  Per  sec. 


Now  for  v  =  10  ft.  per  sec.  and  d  =  %  ft.,  we  find  (in  table, 
§  517)  /=  .00549  ;  and  hence,  from  eq.  (3), 


+  =  12-23  ft, 

of  which  total  head,  as  we  may  call  it,  1.60  ft.  is  used  in  pro- 
ducing the  velocity  10.18  ft.  per  sec.  (i.e.,  vm*  -±-  2g  =  1.60  ft.), 

while  0.808  ft.  (  =  dz^M  is  lost  at  the  entrance  ^(with  a  = 

90°),  and  9.82  ft.  (friction-head)  is  lost  in  skin-friction. 

EXAMPLE  2.  —  [Data  from  Weisbach.]  Required  the  de- 
livery, Q,  through  a  straight  clean  iron  pipe  48  ft.  long  and 
2  in.  in  diameter,  with  5  ft.  head  (=  fi).  V,  =  vm,  being  un- 
known, we  first  takey=  .006  and  obtain  [eq.  (4)] 


Vm=      I  .    .    _    .   4x. 006X48^ 


1  _L.  .505  -f 


=  6.18  ft.  per  sec. 

From  the  table,  §  5 IT,  f or  v  =  6.2  ft.  per  sec.  and  d  =  2  in., 
/=  .00638,  whence 


/ 


Vi+^5+*x-°Tx4sV 

^  7T 

=  6.04  ft.  per  sec., 
which  is  sufficiently  close.     Then,  for  the  volume  per  second, 

Q  =  -  d*vm  —  J7r(^)26.04  =  0.1307  cub.  ft.  per  sec. 
4 


714  MECHANICS   OF   ENGINEERING. 

[Weisbach's  results  in  this  example  are 
vm  =  6.52  ft.  per  sec. 
and  Q  —  0.1420  cub.  ft.  per  sec., 

but  his  values  for/"  are  slightly  different.] 

EXAMPLE  3. — [Data  from  Weisbach.]  What  must  be  the 
diameter  of  a  straight  clean  iron  pipe  100  ft.  in  length,  which 
is  to  deliver  Q  =  \  of  a  cubic  foot  of  water  per  second  under 
5ft.  head  (=  A)  I 

With/^  .006  (approximately),  we  have  from  eq.  (7),  put- 
ting d  —  0  under  the  radical  for  a  first  trial  (ft.,  lb.,  sec.), 


4:0 

whence  v  =  -~  =  7  ft.  per  sec. 

net* 

For  d  =  3.6  in.  and  v  =  7  ft.  per  sec.,  we  find/=  .00601  ; 
whence,  again, 

-,       5  /1.505  x  .30  +  4  X  .00601  X  100     /4  X  JV     n  OCM*  . 

=  V~         2  x  32.2  X  5        ";rTv  = 

and  the  corresponding  v  =  6.06  ft. 

For  this  d  and  v  we  find/=  .00609,  whence,  finally, 

d=  .  7005  x  .30  +  4x  .00609  x  100  /2V  ft_ 

y  2  X  32.2  X  5  \nl 

[Weisbach's  result  is  d  —  .318  ft] 

519.  Ch6zy's  Formula.  —  If,  in  the  problem  of  the  preceding 
paragraph,  the  pipe  is  so  long,  and  therefore  I  :  d  so  great, 
that  4/1  -r-  d  in  eq.  (3)  is  very  large  compared  with  1  +  C#, 
we  may  neglect  the  latter  term  without  appreciable  error; 
whence  eq.  (3)  reduces  to 


ig.  573),     .     .     (8) 


CHEZY'S  FOBMULA.  715 


which  is  known  as  Chezy's  Formula.     For  example,  if  Z  =  100 
ft.  and  d  —  2  in.  =  \  ft,  and/approx.  =  .006,  we  have  A/  -7-  = 

144,  while  1  +  C#  for  square  corners  •=  1.505  only. 
If  in  (8)  we  substitute 


(8)  reduces  to 

.A  =  _-    /—.|L     .     .     .     (very  long  pipe):,     ....     (9) 
7t        01     Ag 

so  that  for  a  very  long  pipe,  considering  f  as  approximately 
constant,  we  may  say  that  to  deliver  a  volume  —  Q  per  unit 
of  time  through  a  pipe  of  given  length  =  I,  the  necessary  head, 
h,  is  inversely  proportional  to  the  ffth  power  of  the  diameter. 

And  again,  solving  (9)  for  Q,  we  find  that  the  volume  con- 
veyed per  unit  of  time  is  directly  proportional  to  the  jift/i power 
of  the  square  root  of  the  diameter  •  directly  proportional  to 
the  square  root  of  the  head  /  and  inversely  proportional  to  the 
square  root  of  the  length.  (Not  true  for  short  pipe ;  see  above 
example.) 

If  we  conceive  of  the  insertion  of  a  great  number  of  piezom- 
eters along  the  long  straight  pipe,  of  uniform  section,  now 
under  consideration,  the  summits  of  the  respective  water 
columns  maintained  in  them  will  lie  in  a  straight  line  joining 
the  discharging  (into  the  air)  end  of  the  pipe  with  a  point  in 
the  reservoir  surface  vertically  over  the  inlet  extremity  (prac- 
tically so),  and  the  "  slope"  of  this  line  (called  the  Hydraulic 
Grade  Line  or  Gradient),  i.e.,  the  tangent  (or  sine ;  the  angle 
is  so  small,  generally)  of  the  angle  which  it  makes  with  the 

horizontal  is  =  — ,  and  may  be  denoted  by  s.     Putting  also 

^d  —  R  =  the  hydraulic  radius  of  the  section  of  the  pipe,  and 
vm  =  v  =  velocity  in  pipe,  we  may  transform  eq.  (8)  into 


or,    v  =  A(B»y,    .    .    .    (9) 


716 


MECHANICS   OF   ENGINEERING. 


which  is  the  form  by  which  Mr.  Hamilton  Smith  (see  §  506) 
interprets  all  the  experiments  quoted  by  him  on  long  pipes. 
As  to  notation,  however,  he  uses  n  for  A,  and  r  for  H.  With 
the  foot  and  second  as  units,  the  quantity  A  (not  an  abstract 
number)  varies  approximately  between  60  and  140.  For  a 
given  A  we  easily  find  the  corresponding  f  from  the  relation 

2# 

y  =  -^ .  If  the  pipe  discharges  under  water,  h  —  the  differ- 
ence of  elevation  of  the  two  reservoirs.  If  the  pipe  is  not 
horizontal,  the  use  of  the  length  of  its  horizontal  projection 

instead  of  its  actual  length  in  the  relation  s  =  —  occasions    an 

i 

error,  but  it  is  in  most  cases  insignificant. 

Similarly,  if  a  steady  flow  is  going  on  in  a  long  pipe  of  uni- 
form section,  at  the  extremities  of  any  portion  of  which  we 
have  measured  the  piezometer  heights  (or  computed  them 
from  the  readings  of  steam  or  pressure  gauges),  we  may  apply 
eq.  (9),  putting  for  A  the  difference  of  level  of  the  piezometer 
summits,  and  for  I  the  length  of  the  pipe  between  them. 

520.  Coefficient  f  in  Fire-engine  Hose. — Mr.  Geo.  A.  Ellis, 
C.E.,  in  his  little  book  on  "  Fire-streams,"  describing  experi- 
ments made  in  Springfield,  Mass.,  gives  a  graphic  comparison 
(p.  45  of  his  book)  of  the  friction-heads  occurring  in  rubber 
hose,  in  leather  hose,  and  in  clean  iron  pipe,  each  of  2j-  in. 
diameter,  with  various  velocities;  on  which  the  following  state- 
merits  may  be  based :  That  for  the  given  size  of  hose  and 
pipe  (d  =  2j-  in.)  the  coefficient  f  for  the  leather  and  rubber 
hose  respectively  may  be  obtained  approximately  by  adding  to 
./for  clean  iron  pipe  (and  a  given  velocity)  the  per  cent  of 
itself  shown  in  the  accompanying  table. 

EXAMPLE. — For  a  clean  iron  pipe 
2^  in.  diam.,  for  a  velocity  =  10  ft. 
per  sec.,  we  have,  from  §  517,  f  = 
.00593.  Hence  for  a  leather  hose  of 
the  same  diameter,  we  have,  for  v  = 
10  ft.  per  sec., 


Velocity 

Rubber 

Leather 

ft.  per 
sec. 

hose  2£  in. 
diam. 

hose  2£  in. 
diam. 

3.0 

50# 

300# 

6.5 

20 

80 

10 

16 

43 

13      • 

12.5 

32 

16 

12 

30 

/=  .00593  +  .43  X  .00593  =  .00848. 


PKESSURE-ENERGY.  717 

521.  Bernoulli's  Theorem  as  an  Expression  of  the  Conservation 
of  Energy  for  the  Liquid  Particles. — In  any  kind  of  flow  with- 
out  friction^  steady  or  not,  in  rigid  immovable  vessels,  the 
aggregate  potential  and  kinetic  energy  of  the  whole  mass  of 
liquid  concerned  is  necessarily  a  constant  quantity  (see  §§  148 
and  149),  but  individual  particles  (as  the  particles  in  the  sink- 
ing free  surface  of  water  in  a  vessel  which  is  rapidly  being 
emptied)  may  be  continually  losing  potential  energy,  i.e., 
reaching  lower  and  lower  levels,  without  any  compensating  in- 
crease of  kinetic  energy  or  of  any  other  kind  ;  but  in  a  steady 
flow  without  friction  in  rigid  motionless  vessels,  we  may  state 
that  the  stock  of  energy  of  a  given  particle,  or  small  collection 
of  particles,  is  constant  during  the  flow,  provided  we  recognize 
a  third  kind  of  energy  which  may  be  called  Pressure-energy, 
or  capacity  for  doing  work  by  virtue  of  internal  fluid  pressure  ; 
as  may  be  thus  explained  : 

In  Fig.  574  let  water,  with  a  very  slow  motion  and  under  a 
pressure  p  (due  to  the  reservoir-head  -f-  atmosphere-head  be- 
hind it),  be  admitted  behind  a  pis- 
ton the    space   beyond  which   is 
vacuous.       Let     s  =  length      of 
I  stroke,  and  F=  the  area  of  pis- 

•    ton.     At  the  end  of  the   stroke, 


f 
^*s^ by  motion  of  proper  valves,  com- 

7/VAC-   munication  with  the  reservoir  is 
FIG-574'  cut  off  on  the  left  of  the  piston 

and  opened  on  the  right,  while  the  water  in  the  cylinder  now  on 
the  left  of  the  piston  is  put  in  'communication  with  the  vacu- 
ous exhaust-chamber.  As  a  consequence  the  internal  pressure 
of  this  water  falls  to  zero  (height  of  cylinder  small),  and  on 
the  return  stroke  is  simply  conveyed  out  of  the  cylinder, 
neither  helping  nor  hindering  the  motion.  That  is,  in  doing 
the  work  of  one  stroke,  viz., 

W  =  force  X  distance  =  Fp  X  s  —  Fps, 

a  volume  of  water  V •=.  Fs,  weighing  Fsy  (Ibs.  or  other  unit)r 
has  been  used,  and,  in  passing  through  the  motor,  has  experi- 
enced* no  appreciable  change  in  velocity  (motion  slow),  and 


718  MECHANICS    OF   ENGINEERING. 

therefore  no  change  in  kinetic  energy,  nor  any  change  of  level, 
and  hence  no  change  in  potential  energy,  but  it  has  given  up 
all  its  pressure.  (See  §  409  for  y.) 

Now  TF,  the  work  obtained  by  the  consumption  of  a  weight 
—  G  =  Vy  of  water,  may  be  written 


.    .    (1) 

Hence  a  weight  of  water  =  G  is  capable  of  doing  the  work 
G  x  —  =  G  X  head  due  to  pressure  p,  i.e.,  —  G  x  pressure- 

head,  in  giving  up  all  its  pressure  p  •  or  otherwise,  while  still 
having  a  pressure^,  a  weight  G  of  water  possesses  an  amount 
of  energy,  which  we  may  call  pressure-energy,  of  an  amount 

£=#•  —  ,  where  y  =  the  heaviness  (§  7)  of  water,  and  -  —  a 

height,  or  head,  measuring  the  pressure  p  ;  i.e.,  it  equals  the 
pressure-head. 

We  may  now  state  Bernoulli's  Theorem  without  friction  in 
a  new  form,  as  follows  :  Multiplying  each  term  of  eq.  (7), 
§  451,  by  Qy,  the  weight  of  water  flowing  per  second  (or  other 
time-unit)  in  the  steady  flow,  we  have 


But  Qy  —  =  -~^-Vm  =  i  X  mass  flowing  per  time  unit  X 

%g     2  9 

square  of  the  velocity  =  the  kinetic  energy  inherent  in  the 
volume  Q  of  water  on  passing  the  section  m,  due  to  the  veloc- 

ity at  m.     Also,  Qy  —  =  the  pressure-energy  of  the  volume 

Q  at  m,  due  to  the  pressure  at  m  ;  while  Qyzm  =  t\iQ  potential 
energy  of  the  volume  Q  at  m  due  to  its  height  zm  above  the 
arbitrary  datum  plane.  Corresponding  statements  may  be 
made  for  the  terms  on  the  right-hand  side  of  (2)  referring  to 
the  other  section,  ??,  of  the  pipe.  Hence  (2)  may  be  thus  read  : 
The  aggregate  amount  of  energy  (of  the  three  kinds  mentioned) 
resident  in  the  particles  of  liquid  when  passing  section  in  is 


LOSS   OF   ENERGY. 


719- 


equal  to  that  when  passing  any  other  section,  as  n  ;  in  steady 
flow  without  friction  in  rigid  motionless  vessels ;  that  is,  the 
store  of  energy  is  constant. 

522.  Bernoulli's  Theorem  with  Friction,  from  the  Standpoint  of 
Energy. — Multiply  each  term  in  the  equation  of  §  516  by  Qy,, 
as  before,  and  denote  a  loss  of  head  or  height  of  resistance  due 
to  any  cause  by  hr ,  and  we  have 


<H  r> 

=  Qrlr+Qy-n 

2g  y 

Each  term  Qyhr(Q.g.,  Qy  4/  -, due  to  skin-friction  in  a 

ff 

long  pipe,  and  Qy  C^^—  due  to  loss  of  head  at  the  reservoir 

entrance  of  a  pipe)  represents  a  loss  of  energy,  occurring  between 
any  position  n  and  any  other  position  m  down-stream  from  % 
but  is  really  still  in  existence  in  the  form  of  heat  generated  by 
the  friction  of  the  liquid  particles  against  each  other  or  the 
sides  of  the  pipes. 

As  illustrative  of  several  points  in  this  connection,  consider 
two  short  lengths  of  pipe  in 
Fig.  575,  A  and  B,  one  offering 
a  gradual,  the  other  a  sudden, 
enlargement  of  section,  but 
otherwise  identical  in  dimen- 
sions. We  suppose  them  to 
occupy  places  in  separate  lines 
of  pipe  in  each  of  which  a 
steady  flow  with  full  cross-sec- 
tions is  proceeding,  and  so  reg- 
ulated that  the  velocity  and  in- 
ternal pressure  at  n,  in  A,  are 
equal  respectively  to  those  at  n 


FIG.  575. 


in  B.     Hence,  if  vacuum  piezometers  be  inserted  at  n,  the- 


720  MECHANICS    OF    ENGINEERING. 

smaller  section,  the  water  columns  maintained  in  them  by  the 

T) 

internal  pressure  will  be  of  the  same  height,  — ,  for  both  A 

and  B.  Since  at  m,  the  larger  section,  the  sectional  area  is  the 
same  for  both  A  and  B,  and  since  Fn  in  A  =*  Fn  in  J5,  so  that 
QA  =  QB>  hence  vm  in  A  =  vm  in  B  and  is  less  than  vn . 

Now  in  B  a  loss  of  head  occurs  (and  hence  a  loss  of  energy) 
between  n  and  ra,  but  none  in  A  (except  slight  friction-head); 
hence  in  A  we  should  find  as  much  energy  present  at  m  as  at 
ny  only  differently  distributed  among  the  three  kinds,  while  at 
m  in  B  the  aggregate  energy  is  less  than  that  at  n  in  B. 

As  regards  kinetic  energy,  there  has  been  a  loss  between  n 
and  m  in  both  A  and  B  (and  equal  losses),  for  vm  is  less  than 
vn .  As  to  potential  energy,  there  is  no  change  between  n  and 
m  either  in  A  or  B,  since  n  and  m  are  on  a  level.  Hence  if 
the  loss  of  kinetic  energy  in  B  is  not  compensated  for  by  an 
equal  gain  of  pressure-energy  (as  it  is  in  A\  the  pressure-head 

(£~\  at  m  in  B  should  be  less  than  that  (S=\    at  rain  A.    Ex- 

\ri*  \y  I  A 

periment  shows  this  to  be  true,  the  loss  of  head  being  due  to 
the  internal  friction  in  the  eddy  occasioned  by  the  sudden  en- 
largement ;  the  water  column  at  m  in  B  is  found  to  be  of  a 
less  height  than  that  at  m  in  A,  whereas  at  n  they  are  equal. 
(See  p.  467  of  article  "  Hydromechanics"  in  the  Ency.  Bri- 
tannica  for  Mr.  Froude's  experiments.) 

In  brief,  in  A  the  loss  of  kinetic  energy  has  been  made  up 
in  pressure-energy,  with  no  change  of  potential  energy,  but  in 
B  there  is  an  actual  absolute  loss  of  energy  =  Qyhr ,  or 

v  a 
=  Qy£  -^-,  suffered  by  the  weight  Qy  of  liquid.     The  value 

? 
of  C  in  this  case  and  others  will  be  considered  in  subsequent 

paragraphs. 

Similarly,  losses  of  head,  and  therefore  losses  of  energy, 
occur  at  elbows,  sharp  bends,  and  obstructions,  causing  eddies 
and  internal  friction,  the  amount  of  each  loss  for  a  given 

weight,  6r,  of  water  being  =  Ghr  =  GC,  ~  ;     Ar  =  C  —  being 

•/  t/ 

the  loss  of  head  occasioned  by  the  obstruction  (§  474).     It  is 


SUDDEN   ENLARGEMENTS  IN   PIPES.  721 

therefore  very  important  in  transmitting  water  through  pipes 
for  purposes  of  power  to  use  all  possible  means  of  preventing 
disturbance  and  eddying  among  the  liquid  particles.  E.g., 
sharp  corners,  turns,  elbows,  abrupt  changes  of  section,  should 
be  avoided  in  the  design  of  the  supply-pipe.  The  amount  of 
the  losses  of  head,  or  heights  of  resistance,  due  to  these  various 
causes  will  now  be  considered  (except  skin-friction,  already 
treated).  Each  such  loss  of  head  will  be  written  in  the  form 

/y2 
C  —  ,  and  we  are  principally  concerned  with  the  value  of  the 

o 
abstract  number  £,  or  coefficient  of  resistance •,  in  each  case. 

The  velocity  v  is  the  velocity,  known  or  unknown,  where  the 
resistance  occurs y  or  if  the  section  of  pipe  changes  at  this 
place,  then  v  =  velocity  on  the  down-stream  section.  The  late 
Professor  "Weisbach,  of  the  mining-school  of  Freiberg,  Saxony, 
was  one  of  the  most  noted  experimenters  in  this  respect,  and 
will  be  frequently  quoted. 

523.  Loss  of  Head  Due  to  Sudden  (i.e.,  Square-edged)  Enlarge- 
ment. Borda's  Formula. — Fig.  576.  An  eddy  is  formed  in  the 
t  angle  with  consequent  loss  of  energy.  Since 
eacn  particle  of  water  of  weight  =  Gl ,  arriving 
with  the  velocity  vl  in  the  small  pipe,  may  be 
t  considered  to  have  an  impact  against  the  base 
.  576.  of  the  infinitely  great  and  more  slowly  moving 
column  of  water  in  the  large  pipe,  and,  after  the  impact, 
moves  on  with  the  same  velocity,  -ya,  as  that  column,  just  as 
occurs  in  inelastic  direct  central  impact  (§  60),  we  may  find 
the  energy  lost  by  this  particle  on  account  of  the  impact  by 
eq.  (1)  of  §  138,  in  which,  putting  Ml  =  6r,  -f-  <?,  and  M^  = 
Gz  -f-  g  =  mass  of  infinitely  great  body  of  water  in  the  large 
pipe,  so  that  M^  —  oo ,  we  have 

Energy  lost  by  particle  =  G1  ^  7  ^-,      .     .     (1) 

2g 

and  the  corresponding 

Loss  of  head  =  -^-L— — --, 


722  MECHANICS   OF  ENGINEERING. 

which,  since  F1v1  =  JF[vt  ,  may  be  written 

\~F         ~~l  3  v  8 
Loss  of  head  in  sudden  enlargement  —    —  f  —  1    —  L. 

That  is,  the  coefficient  C  for  a  sudden  enlargement  is 


(2) 


.     .    .    .    .     .     (3) 

Fl  and  Fa  are  the  respective  sectional  areas  of  the  pipes.     Eq. 
(2)  is  Bordtis  Formula. 

NOTE.  —  Practically,  the  flow  cannot  always  be  maintained 
with  full  sections.  In  any  case,  if  we  assume  the  pipes  to  be 
running  full  (once  started  so),  and  on  that  assumption  compute 
the  internal  pressure  at  F1  ,  and  find  it  to  be  zero  or  negative, 
the  assumption  is  incorrect.  That  is,  unless  there  is  some 
pressure  at  Fl  the  water  will  not  swell  out  laterally  to  fill  the 
large  pipe. 

EXAMPLE.  —  Fig.  577.  In  the  short  tube  AB  containing  a 
sudden  enlargement,  we  have  given  F9  —  Fm  —  6  sq.  inches, 
FI  =  4  sq.  inches,  and  h  —  9  feet.  Re- 
quired  the  velocity  of  the  jet  at  m  (in 
the  air,  so  t\ia.tpm  -f-  y  =  1}  —  34  ft.),  if 
the  only  loss  of  head  considered  is  that 
due  to  the  sudden  enlargement  (skin- 
friction  neglected,  as  the  tube  is  short  ; 
the  reservoir  entrance  has  rounded  cor- 
ners).  Applying  Bernoulli's  Theorem 
to  m  as  down-stream  section,  and  n  in  reservoir  surface  as  up- 
stream position  (datum  level  at  w),  we  have 


FIO.  577. 


But,  here,  v,  =  vm; 


From  eq.  (3)  we  have 


(5) 


C  =  (*  -  I)1  =  0.25, 


SUDDEN  ENLARGEMENT  IN  PIPE.  723 

and  finally  (ft.,  lb.,  sec.) 


v*=\      r      1/2  X  32.2  X  9  =  0.895  V2X  32.2x9 
W   1.^5 

=  21.55  ft.  per  sec. 

(The  factor  0.895  might  be  called  a  coefficient  of  velocity  for 
this  case.)     Hence  the  volume  of  flow  per  second  is 

Q  =  Fmvm  =  -ThX  21.55  =  0.898  cub.  ft.  per  sec. 

We  have  so  far  assumed  that  the  water  fills  both  parts  of  the 
tube,  i.e.,  that  the  pressure  J915  at  F^  ,  is  greater  than  zero  (see 
foregoing  note).  To  verify  this  assumption,  we  compute  p^ 
by  applying  Bernoulli's  Theorem  to  the  centre  of  F^  as  down- 
stream position  and  datum  plane,  and  n  as  up-stream  position, 
with  no  loss  of  head  between,  and  obtain 

j£  +  7  +  0  =  °  +  ft  +  A-°-  •  -  -  •  (6> 

But  since  Fp^  =  F^v^  ,  we  have 


<  =  (*)v  -  a  )x°, 

and  hence  the  pressure-head  at  F,  (substituting  from  equations 
above)  is 


and  /.  pi  =  ||-  of  14.7  =  11.6  Ibs.  per  sq.  inch,  which  is 
greater  than  zero  ;  hence  efflux  with  the  tube  full  in  both  parts 
can  be  maintained  under  9  ft.  head. 

If,  with  Fl  and  F^  as  before  (and  .'.  C),  we  put  p,  =  0,  and 
solve  for  7i,  we  obtain  h  =  42.5  ft.  as  the  maximum  head 
under  which  efflux  with  the  large  portion  full  can  be  secured. 

524.  Short  Pipe,  Square-edged  Internally.  —  This  case,  already 


724  MECHANICS    OF   ENGINEERING. 

treated  in  §§  507  and  515  (see  Fig.  578 ;  a  repetition  of  560), 
presents  a  loss  of  head  due  to  the  sudden  enlargement  from 
the  contracted  section  at  mf  (whose  sec- 
*  tional  area  may  be  put  —  CF,  C  being 
i  an  unknown  coefficient,  or  ratio,  of 
j  contraction)  to  the  full  section  F  of 
:'_l:r  the  pipe.  From  §  515  we  know  that 
^j=^=.  the  loss  of  head  due  to  the  short  pipe 

is  hr  =  CJT^  (for  a  =  90°),  in  which 
FIG.  578.  CE  =  0.505  ;  while  from  Borda's  For- 

mula, §  523,  we  have  also  £E  =    —^  —  1     .    Equating  these, 
we  find  the  coefficient  of  internal  contraction  at  mf  to  be 


C=  -  -=  =  -  -  --  =  0.584, 


or  about  0.60  (compare  with  C=  .64  for  thin-plate  contrac- 
tion, §  497).  It  is  probably  somewhat  larger  than  this  (.584), 
since  a  small  part  of  the  loss  of  head,  Ar,  is  due  to  friction  at 
the  corners  and  against  the  sides  of  the  pipe. 

By  a  method  similar  to  that  pursued  in  the  example  of 
§  523,  we  may  show  that  unless  h  is  less  than  40  feet,  about, 
the  tube  cannot  be  kept  full,  the  discharge  being  as  in  Fig. 
551.  If  the  efflux  takes  place  into  a  "  partial  vacuum,"  this 
limiting  value  of  h  is  still  smaller.  "Weisbach's  experiments 
confirm  these  statements  (but  those  in  the  C.  U.  Hyd.  Lab. 
seem  to  indicate  that  the  limiting  value  for  h  in  the  first  case 
is  about  50  ft.). 

525.  Diaphragm  in  a  Cylindrical  Pipe.—  Fig.  579.  The  dia- 
phragm being  of  "thin  plate,"  , 

let  the  circular  opening  in  it 
(concentric  with  the  pipe)  have 
an  area  =  F,  while  the  sectional 
area  of  pipe  =  F9  .  Beyond  F,  the  FIG.  579. 

stream  contracts  to  a  section  of  area  =  CF  —  Fl  ,  in  enlarging 


SHOKT   PIPE.       DIAPHRAGM 


PIPE. 


725 


-from  which  to  fill  the  section  F^  ,  of  pipe,  a  loss  of  head  occurs 
which  by  Borda's  Formula,  §  523,  is 


F1 

--         *        1 

-- 


where  v2  is  the  velocity  in  the  pipe  (supposed full).  Of  course 
Fl  (or  OF)  depends  on  F\  but  since  experiments  are  necessary 
in  any  event,  it  is  just  as  well  to  give  the  values  of  £  itself,  as 
determined  by  Weisbach's  experiments,  viz. : 


For~-  =  .10 

#1 

.20 

.30 

.40 

.50 

.60 

.70 

.80 

.90 

1.00 

C  =  226. 

48. 

17.5 

7.8 

3.7 

1.8 

.8 

.3 

.06 

0.00 

By  internal  lateral  filling,  Fig.  580,  the  change  of  section 
may  be  made  gradual  and  eddying 
prevented  ;  and  then  but  little  loss 
of  head  (and  therefore  little  loss  of 
energy)  occurs,  besides  the  slight 
amount  due  to  skin-friction  along 
this  small  surface. 


FIG.  580. 

On  p.  467  of  the  article  Hydromechanics 
in  the  Encyclopaedia  Britannica  may  be  found  an  account  of 
experiments  by  Mr.  Froude,  illustrating  this  fact. 


526.  "The  Venturi  Water-meter." — The  invention  bearing 
this  name  was  made  by  Mr.  Clemens  Herschel  (see  Trans.  Am- 
Soc.  Civ.  Engineers,  for  November  1887),  and  may  be  de- 
scribed as  a  portion  of  pipe  in  which  a  gradual  narrowing  of 
section  is  immediately  succeeded  by  a  more  gradual  enlarge- 
ment, as  in  Fig.  580 ;  but  the  dimensions  are  more  extreme. 
During  the  flow  the  piezometer-heights  are  observed  at  the 
three  positions  r,  n,  and  m  (see  below),  and  the  rate  of  dis- 
charge may  be  computed  as  follows :  Referring  to  Fig.  580, 
let  us  denote  by  r  the  (up-stream)  position  where  the  narrow- 
ing of  the  pipe  begins,  and  by  m  that  where  the  enlargement 
ends,  while  n  refers  to  the  narrowest  section.  Fm  =  Fr. 

Applying  Bernoulli's  Theorem  to  sections  r  and  n,  assuming 


726  MECHANICS    OF   ENGINEERING. 

no  loss  of  head  between,  we  have,  as  the  principle  of  the  ap- 
paratus, 


. 


whence,  since  Frvr  =  Fnvn  , 


» 


in  which  0  represents  the  first  radical  factor.     0  should  differ 

F 

but  little  from  unity  with  -£  small  (and  such  was  found  to  be 

/V 

the  case  by  experiment).  Its  theoretical  value  is  constant  and 
greater  than  unity.  In  the  actual  use  of  the  instrument  the 

—  and  •&  are  inferred  from  the  observed  piezometer-heights 

yr  and  yn  (since  &  =  yr  +  1,  and  ^  =  yn  +  &,  5  being  =  34  ft.), 

and  then  the  quantity  flowing  per  time-unit  computed,  from 
Q  —  Fnvn,  vn  having  been  obtained  from  eq.  (2).  This  pro- 
cess gives  a  value  of  Q  about  four  per  cent  in  excess  of  the 
truth,  according  to  the  second  set  of  experiments  mentioned 
below,  if  vn  =35  ft.  per  sec.  ;  but  only  one  per  cent  excess  with 
vn  =  5  or  6  ft.  per  sec. 

Experiments  were  made  by  Mr.  Herschel  on  two  meters  of 
this  kind,  in  each  of  which  Fn  was  only  one  ninth  of  Fr  ,  a 
ratio  so  extreme  that  the  loss  of  head  due  to  passage  through 
the  instrument  is  considerable.  E.g.,  with  the  smaller  appara- 
tus, in  which  the  diameter  at  n  was  4  in.,  the  loss  of  head  be- 
tween r  and  in  was  10  or  11  ft.,  when  the  velocity  through  n 
was  50  ft.  per  sec.,  those  at  other  velocities  being  roughly  pro- 
portional to  the  square  of  the  velocity.  In  the  larger  instru- 
ment dn  was  3  ft.,  and  the  loss  of  head  between  r  and  m  was 
much  more  nearly  proportional  to  the  square  of  the  velocity 
than  in  the  smaller.  (E.g.,  with  vn  =  34.56  ft.  per  sec.  the 
loss  of  head  was  2.07  ft.,  while  with  vn  =  16.96  ft.  per  sec.  it 


SUDDEN  DIMINUTION   OF   SECTION   OF  PIPE. 


727 


was  0.49  ft.)     The  angle  of  divergence  was  much  smaller  in 
these  meters  than  that  in  Fig.  580. 

527.  Sudden  Diminution  of  Cross-section,  Square  Edges. — Fig. 
581.     Here,  again,  the  resistance  is  ^ 

due  to  the  sudden  enlargement  from     =^  f^^\^_  _____ ^ 

the  contracted  section  to  the  full  sec-  -^.-J^&j 
tion  F^  of  the  small  pipe,  so  that  in  ^~~~-^3 
the  loss  of  head,  by  Borda's  formula, 


FIG.  581. 


the  coefficient 


depends  on  the  coefficient  of  contraction  C\  but  this  latter  is 
influenced  by  the  ratio  of  F^  to  F0,  the  sectional  area  of  the 
larger  pipe,  C  being  about  .60  when  F0  is  very  large  (i.e., 
when  the  small  pipe  issues  directly  from  a  large  reservoir  so 
that  Fz  :  F0  practically  =  0).  For  other  values  of  this  ratio 
TVeisbach  gives  the  following  table  for  (7,  from  his  own  ex- 
periments : 


For  F*  :  F0  =  .10 

.20 

.30 

.40 

.50 

.60 

.70 

.80 

.90 

1.00 

C=  .624 

.632 

.643 

.659 

.681 

.712 

.755 

.813 

.892 

1.00 

C   being  found,   we   compute    C    from  eq.   (2)   for  use  in 

eq.  (1). 

528.  Elbows. — The  internal  disturbance  caused  by  an  elbow, 
Fig.  582  (pipe  full,  both  sides  of  elbow),  occasions  a  loss  of 
head 


FIG.  582.  in  which,  according  to  Weisbach's  experi- 

ments with  tubes  3  centims.,  i.e.  1.2  in.,  in  diameter,  we  may 
put 


728 


MECHANICS    OF   ENGINEERING. 


For  a  -    20° 

40° 

60° 

8(T 

90° 

100° 

110° 

120° 

130° 

140° 

C=  .046 

.139 

.364 

.740 

.984 

1.26 

1.556 

1.86 

2.16 

2.43 

computed  from  the  empirical  formula  ; 

C  =  .9457  sin'  %a  +  2.047  sin4  £« ; 

v  is  the  velocity  in  pipe ;  a  as  in  figure.     For  larger  pipes  C 
would  probably  be  somewhat  smaller ;  and  vice  versa. 

If  the  elbow  is  immediately  succeeded  by  another  in  the  same 
plane  and  turning  the  same  way,  Fig.  583,  the 
loss  of  head  is  not  materially  increased,  since 
the  eddying  takes  place  chiefly 
in  the  further  branch  of  the 
second  elbow ;  but  if  it  turns 
in  the  reverse  direction,  Fig. 
584,  but  still  in  the  same 
plane,  the  total  loss  of  head  is  double  that  of 
one  elbow ;  while  if  the  plane  of  the  second  is  ~| 
to  that  of  the  first,  the  total  loss  of  head  is  1-J- 
times  that  of  one  alone.  (Weisbach.) 


FIG.  583. 


FIG.  584. 


529.  Bends  in  Pipes  of  Circular  Section. — Fig.  585.  Weis- 
bach bases  the  following  empirical 
formula  for  C,  the  coefficient  of  resist- 
ange  of  a  quadrant  bend  in  a  pipe 
of  circular  section,  on  his  own  experi- 
FIG.  585.  ments  and  some  of  Dubuat's,  viz. : 


C  =  0.131  + 1.847 


for  use  in 


(1> 

(2) 


where  a  =  radius  of  pipe,  r  —  radius  of  bend  (to  centre  of 
pipe),  and  v  —  velocity  in  pipe ;  hr  =  loss  of  head  due  to 
bend. 


ELBOWS   AND   BENDS   IN   PIPES. 


729 


It  is  understood  that  the  portion  EC  of  the  pipe  is  kept  full 
by  the  flow ;  which,  however,  may  not  be  practicable  unless 
BG\§  more  than  three  or  four  times  as  long  as 
wide,  and  is  full  at  the   outset.     A  semicircular 
bend  occasions  about  the  same  loss  of  head  as  a 
quadrant  bend,  but  two  quadrants  forming  a  re- 
verse curve  in  the  same  plane,  Fig.  586,  occasion  a 
double  loss.     By  enlarging  the   pipe  at  the  bend, 
or  providing  internal  thin  partitions  parallel  to  the 
sides,  the  loss  of  head  may  be  considerably  dimin- 
ished.    Weisbach  gives   the  following  table  com-      FIG.  586. 
puted  from  eq.  (1),  but  does  not  state  the  absolute  size  of  the 
pipes. 


For  -=.10 
r 

.20 

.30 

.40 

.50 

.60 

.70 

.80 

.90 

1.0 

C=.131 

.138 

.158 

.206 

.294 

.440 

.661 

.977 

1.40 

1.98 

Accounts  of  many  of  Weisbach's  hydraulic  experiments  are 
contained  in  the  Civilingenieur,  vols.  ix,  x,  and  xi. 

529a.  Common  Pipe-elbows. — Prof.  L.  F.  Bellinger  of  Nor- 
wich University,  Vermont,  conducted  a  set  of  experiments  in 
1887,  when  a  student  at  Cornell,  on  the 
loss  of  head  occasioned  by  a  common  el- 
bow (for  wrought-iron  pipe),  whose  longi- 
tudinal section  is  shown  in  Fig.  586a. 
The  elbow  served  to  connect  at  right 
angles  two  wrought-iron  pipes  having  an 
internal  diameter  of  0.482  in. 

The  internal  diameter  of  the  short  bend 
or  elbow  was  f  in.,  and  the  radius  of  its  curved  circular  axis 
(a  quadrant)  was  £  in.  Its  internal  surface  was  that  of  an 
ordinary  rough  casting. 

A  straight  pipe  of  the  same  character  and  size  and  14  feet 
long  was  first  used,  and  the  loss  of  head  due  to  skin-friction 
(the  only  loss  of  head  in  that  case)  carefully  determined  for  a 
range  of  velocities  from  2  to  20  ft.  per  sec. 

Two  lengths  of  similar  pipe  were  then  joined  by  the  elbow 


FIG.  586a. 


730 


MECHANICS   OF   ENGINEERING. 


mentioned,  forming  a  total  length  of  14  feet,  and  the  total  loss 
of  head  again  determined  through  the  same  range  of  velocities. 
By  subtraction,  the  loss  of  head  due  to  the  elbow  was  then 
easily  found  for  each  velocity,  and  assuming  the  form 


(1) 


for  the  loss  of  head,  C  was  computed  in  each  case. 

From  Fig.  586a  it  is  seen  that  the  stream  meets  with  a  sud- 
den enlargement  and  a  sudden  diminution,  of  section,  as  well 
as  with  the  short  bend ;  so  that  the  disturbance  is  of  a  rather 
complex  nature. 

The  principal  results  of  Prof.  Bellinger's  experiments,  after 
the  adjustment  of  the  observed  quantities  by  "  least  squares," 
were  found  capable  of  being  represented  fairly  well  by  the 
formula 

C  =  0.621  +  [2W  -  1]  X  0.0376,       ..   .  '  .     (2) 

where  n  =  [veloc.  in  pipe  in  ft.  per  sec.]  -r-  5.     The  following 
table  was  computed  from  eq.  (2)  (where  v  is  in  ft.  per  second) : 


V  = 

2 

4 

6 

8 

10 

12 

14 

16 

18 

20 

&=• 

.633 

.649 

.670 

.697 

.734 

.782 

.845 

.929 

1.039 

1.185 

530.  Valve-gates  and  Throttle-valves  in  Cylindrical  Pipes. — 
Adopting,  as  usual,  the  form 

*-'t|p  •  ......  a) 

for  the  loss  of  head  due  to  a  valve-gate,  Fig.  587,  or  for  a 

find 

,^r^        r_  i____J  1 

nm 


^Sf^jt       ~\  ~  ^>>  WX7<^ 


FIG.  587.  FIG.  588. 

throttle-valve,  Fig.  588,  each  in  a  definite  position,  Weisbach's 


VALVE-GATE.      THROTTLE-VALVE. 


731 


•experiments  furnish  us  with  a  range  of  values  of  C  in  the  case 
of  these  obstacles  in  a  cylindrical  pipe  1.6  inches  in  diameter, 
:as  follows  (for  meaning  of  s,  d,  and  a,  see  figures,  v  is 
the  velocity  in  the  full  section  of  pipe,  running  full  on  both 
.sides.) 


Valve-gate. 

Throttle-valve. 

S 

d 

£ 

a 

•9 

5° 

.24 

1.0 

.00 

10° 

.52 

1 

.07 

15° 

20° 

.90 
1.54 

1 

.26 

25° 
30° 

2.51 
3.91 

* 

.81 

35° 

6.22 

40° 

10.8 

1 

2.06 

45° 

18.7 

1 

5.52 

50° 
55° 

32.6 

58.8 

I 

17.00 

60° 
65° 

118.0 
256.0 

* 

97.8 

70° 

751. 

531.  Examples  involving  Divers  Losses  of  Head. — We  here 
suppose,  as  before,  that  the  pipes  are  full  during  the  flow. 
Practically,  provision  must  be  made  for  the  escape  of  the  air 
which  collects  at  the  high  points.  If  this  air  is  at  a  tension 
greater  than  one  atmosphere,  automatic  air-valves  will  serve  to 
provide  for  its  escape ;  if  less  than  one  atmosphere,  an  air- 
pump  can  be  used,  as  in  the  case  of  a  siphon  used  at  the 
Kansas  City  Water  Works.  (See  p.  346  of  the  Engineering 
News  for  November  1887.) 


:'"•;..',;.  »;•  .." 

m 

X    90° 

...                       -,--              -^ 
7n=  50 

—  _rr-r^ 

~-\vA 

A 

0         f 

0                               7        on 

—      ' 

^E=^='- 

:•) 

n 

—                . 

— 

S 

I 

90°                  K                          £fei'V 

l 

1 

To  "         & 

FIG.  589. 


EXAMPLE  1. — Fig.  589.     What  head,  =  A,  will  be  required 
to  deliver  \  U.  S.  gallon  (i.e.  231  cubic  inches)  per  second 


732  MECHANICS   OF  ENGINEEKING. 

through  the  continuous  line  of  pipe  in  the  figure,  containing  two 
sizes  of  cylindrical  pipe  (d0  =  3  in.,  and  d2  =  1  in.),  and  two 
90°  elbows  in  the  larger.  The  flow  is  into  the  air  at  ra,  the 
jet  being  1  in.  in  diameter,  like  the  pipe.  At  E,  a  =  90°,  and 
the  corners  are  not  rounded  ;  at  JT,  also,  corners  not  rounded. 
Use  the  ft.-lb.-sec.  system  of  units  in  which  g  =  32.2. 

Since  Q  =  %  gal.  =  j-  •  T2TyB  =  .0668  cub.  ft.  per  sec.,  and 
therefore  the  velocity  of  the  jet 

vm  =  v,  =  Q  -f-  i?^)2  =  12.25  ft.  per  sec.; 

hence  the  velocity  in  the  large  pipe  is  to  be  v0  =  (-i-)X  =  1.36 
ft.  per  sec.  From  Bernoulli's  Theorem,  we  have,  with  m  as 
datum  plane, 


involving  six  separate  losses  of  head,  for  each  of  which  there 
is  no  difficulty  in  finding  the  proper  £  or/*,  since  the  velocities 
and  dimensions  are  all  known,  by  consulting  preceding  para- 
graphs. (Clean  iron  pipe.) 

From  §  515,  table,  for  a  =  90°  we  have  .    ./'  ,.     C^  =  0.505 

"     §51Vor<=3in.,and^0=1.36ft.persec.,/0=:    -00725 
"          "       "  d,  =  l  in.,  and  v9  -=12.25  "     "       /2  =   .00613 
*     §  528  (elbows),  for  «  =  90°      ....     Cel  =0.984 
"     §  527,  for  sudden  diminution  at  K  we  have 
[since  F^  +  F0  =  I2  ~  32  =  0.111,  /.  (7  =  0.625] 


Solving  the  above  equation  for  A,  then,  and  substituting 
above  numerical  values  (in  ft.-lb.-scc.-system),  we  have  (noting 
that  vm  '=  v»  and  v0  =  ^2) 


(.606  + 
» 


9^A  ,  4  X.  00613X201. 
.360  -|  --  -  U 


EXAMPLES;    WITH  LOSSES  OF  HEAD. 


733 


i.e., 


h  =  ^1^  [  1  +(.00623  +.07160  +  .0243)+(.360+5.8848~]; 

t)4:.4:        L_ 

.-.  h  =  2.323  X  7.3469  =  17.09  ft.—Ans. 

It  is  here  noticeable  how  small  are  the  losses  of  head  in  the 
large  pipe,  the  principal  reason  of  this  being  that  the  velocity 
in  it  is  so  small  (v0  =  only  1.36  ft.  per  sec.),  and  that  in  gen- 
eral losses  of  head  depend  on  the  square  of  the  velocity 
(nearly). 

In  other  words,  the  large  pipe  approximates  to  being  a  reser- 
voir in  itself. 

With  no  resistances  a  head  h  =  0OTa  -~  2g  =  2.32  ft.  would  be 
sufficient. 

EXAMPLE  2. — Fig.  590.  "With  the  valve-gate  Thalf  raised 
(i.e.,  8  =  %d  in  Fig.  587),  required  the  volume  delivered  per 
second  through  the  clean  pipe  here  shown.  The  jet  issues 


FIG.  590. 


from  a  short  straight  pipe,  or  nozzle  (of  diameter  d^  =  1£  in.) 
inserted  in  the  end  of  the  larger  pipe,  with  the  inner  corners 
not  rounded.  Dimensions  as  in  figure.  Hadius  of  each  bend 
=  r  =  2  in.  The  velocity  vm  of  the  jet  in  the  air  =  velocity 
i>2  in  the  small  pipe ;  hence  the  loss  of  head  at 


Now  vm  is  unknown,  as  yet ;  but  VQ  ,  the  velocity  in  the  large 
pipe,  is  =  vm  ntF]>  i-e-J  v°  =  A1"*-     From  Bernoulli's  The- 


734  MECHANICS   OF   ENGINEERING. 

•orem  (m  as  datum  level)  we  obtain,  after  transposition, 


Of  the  coefficients  concerned,  f0  alone  depends  on  the  un- 
known velocity  v0.     For  the  present  [first  approximation], 

put /0  =  .006 

From  §  515,  with  a  =  90°,  .     .......    CE  =  .505 

From  §  517,  valve-gate  with  s  =  %d, CF  =  2.06 

From  §  529,  with  a:r  =  0.5, CB  =  0.294 

While  at  K,  from  §  527,  having 

(F,  :  F0)  =  (f)2  :  22  =  -&  =  0.562; 

we  find  from  table,      .     .     .    ,     .    . '  .    .     .     .     C  =  0.700 
and  .-.  c*=(J_-l)2=(0.428)2.     .     .     .    i.e.,  £*=  0.183 

Substituting  in  eq.  (1)  above,  with  v*  =  (TV)2^m2>  we  have 

'  •  -, » 


in  which  the  first  radical,  an  abstract  number,  might  be  called 
a  coefficient  of  velocity,  0,  for  the  whole  delivery  pipe ;  and 
also,  since   in   this   case   Q,  =  Fmvm  =  Fjvz ,  may  be  written 
Q  =  }JiF^  V^g/iy  it  may  be  named  a  coefficient  of  efflux,  p. 
Hence 


•  • 


V2  x  32.2  x  25: 


A  vm=  0.421  i/2^A  =  0.421  ^2  X  32.2  X  25  =  16.89  ft.  per  sec, 

(The  .421  might  be   called  a   coefficient  of  velocity.)     The 
volume  delivered  per  second  is 

Q  =  \nd;  vm  =  i?r(A)3  16-89  =  -2°7  cu^.  ft.  per  sec. 


(As  the  section  of  the  jet  Fm  —  F^  ,  that  of  the  short  pipe  or 
nozzle,  we  might  also  say  that  .421  =  /*  =  coefficient  of  efflux, 
for  we  may  write  Q  =  pF^  V'2gh,  whence  /*  —  .421.) 


FLOW   THROUGH   SIPHONS.  735 

532.  Siphons.— In  Fig.  532,  §  490,  the  portion  HNJ)  is 
above  the  level,  BC>  of  the  surface  of  the  water  in  the  head 
reservoir  BL,  and  yet  under  proper  conditions  a  steady  flow 
can  be  maintained  with  all  parts  of  the  pipe  full  of  water,  in- 
cluding HJV^O.  If  the  atmosphere  exerted  no  pressure,  this 
would  be  impossible ;  but  its  average  tension  of  14.7  Ibs.  per 
sq.  inch  is  equivalent  to  an  additional  depth  of  nearly  34  feet 
of  water  placed  upon  BC.  With  no  flow,  or  a  very  small 
velocity,  the  pipe  may  be  kept  full  if  JVZ  is  not  more  than 
33  or  34  feet  above  BC\  but  the  greater  -ya,  the  velocity  of 
flow  at  NI,  and  the  greater  and  more  numerous  the  losses 
of  head  between  L  and  N^  the  less  must  be  the  height  of  N^ 
above  BC  for  a  steady  flow. 

The  analytical  criterion  as  to  whether  a  flow  can  be  main- 
tained or  not,  supposing  the  pipe  completely  filled  at  the  out- 
set, is  that  the  internal  pressure  must  be  >  0  at  all  parts  of 
the  pipe.  If  on  the  supposition  of  a  flow  through  a  pipe  of 
given  design  the  pressure^?  is  found  <  0,  i.e.  negative,  at  any 
point  \N^  being  the  important  section  for  test]  the  supposition 
is  inadmissible,  and  the  design  must  be  altered. 

For  example,  Fig.  532,  suppose  LN^N^  to  be  a  long  pipe  of 
uniform  section  (diameter  =  d,  and  length  =  Z),  and  that  under 
the  assumption  of  filled  sections  we  have  computed  i>4,  the 
velocity  of  the  jet  at  ^4 ;  i.e., 


To  test  the  supposition,  apply  Bernoulli's  Theorem  to  the 
surface  BC  and  the  point  N^  where  the  pressure  is  ^>2,  velocity 
•ya(=  -y0  since  we  have  supposed  a  uniform  section  for  whole 
pipe),  and  height  above  BC=h^.  Also,  let  length  of  pipe 
=  Za  .  Whence  we  have 


[BC  being  datum,  plane.] 


736  MECHANICS   OF   ENGINEERING. 

Solving  for  —  ,  we  have 

!'  =  34  feet  -  [A,  +  g-  (l  +  Cji+  4/|].  .    .    (3) 
We  note,  then,  that  for  j?a  to  be  <  0, 

---^f1-..   (4) 


In  the  practical  working  of  a  siphon  it  is  found  that  atmos- 
pheric air,  previously  dissolved  in  the  water,  gradually  collects 
at  N^  ,  the  highest  point,  during  the  flow  and  finally,  if  not  re- 
moved, causes  the  latter  to  cease.  See  reference  below. 

One  device  for  removing  the  air  consists  in  first  allowing  it 
to  collect  in  a  chamber  in  communication  with  the  pipe  be- 
neath. This  communication  is  closed  by  a  stop-cock  after  the 
water  in  it  has  been  completely  displaced  by  air.  Another 
stop-cock,  above,  being  now  opened,  water  is  poured  in  to  re- 
place the  air,  which  now  escapes.  Then  the  upper  stop-cock  is 
shut  and  the  lower  one  opened.  The  same  operation  is  again 
necessary,  after  some  hours. 

On  p.  346  of  the  Engineering  News  of  November  1887  may 
be  found  an  account  of  a  siphon  which  has  been  employed  since 
1875  in  connection  with  the  water-works  at  Kansas  City.  It 
is  1350  ft.  long,  and  transmits  water  from  the  river  to  the 
artificial  "  well  "  from  which  the  pumping  engines  draw  their 
supply.  At  the  highest  point,  which  is  16  ft.  above  low-  water 
level  of  the  river,  is  placed  a  "  vacuum  chamber  "  in  which  the 
air  collects  under  a  low  tension  corresponding  to  the  height, 
and  a  pump  is  kept  constantly  at  work  to  remove  the  air  and 
prevent  the  "  breaking"  of  the  (partial)  vacuum.  The  diam- 
eter of  the  pipe  is  24  in.,  and  the  extremity  in  the  "  well  "  dips 
5  ft.  below  the  level  of  low  water.  See  Trautwine's  Pocket- 
book,  for  an  account  of  Maj.  Crozet's  Siphon. 

532a.  Branching  Pipes.  —  If  the  flow  of  water  in  a  pipe  is 
caused  to  divide  and  pass  into  two  others  having  a  common 


BRANCHING   PIPES.  737 

junction  with  the  first,  or  vice  versa  ;  or  if  lateral  pipes  lead 
out  of  a  main  pipe,  the  problem  presented  may  be  very  com- 
plicated. As  a  comparatively  simple  instance,  let  us  suppose 
that  a  pipe  of  diameter  d  and  length  I  leads  out  of  a  reservoir, 
and  at  its  extremity  is  joined  to  two  others  of  diameters  dl  and 
d^  and  lengths  lt  and  Z2  respectively,  and  that  the  further  extrem- 
ities of  the  latter  discharge  into  the  air  without  nozzles  under 
heads  Ax  and  A2  below  the  reservoir  surface.  Call  these  two 
pipes  Nos.  1  and  2.  That  is,  the  system  forms  a  Y  in  plan. 

Assuming  that  all  entrances  and  junctions  are  smoothly 
rounded,  so  that  all  loss  of  head  is  due  to  skin-friction,  it  is  re- 
quired to  iind  the  three  velocities  of  flow,  v,  vl9  and  va,  in  the 
respective  pipes.  First  applying  Bernoulli's  Theorem  to  a 
stream-line  from  the  reservoir  surface  through  the  main  pipe 
to  the  jet  at  the  discharging  end  of  pipe  No.  1,  we  have 


and  similarly,  dealing  with  a  stream-line  through 
pipe  and  No.  2, 


while  the  equation  of  continuity  for  this  case  is 

.      ....     (3) 


From  these  three  equations,  assuming/'  the  same  in  all  pipes 
as  a  first  approximation,  we  can  find  the  three  velocities  (best 
by  numerical  trial,  perhaps)  ;  and  then  the  volume  of  discharge 
of  the  system  per  unit  of  time 


(4) 


533.  Time  of  Emptying  Vertical  Prismatic  Vessels  (or  Inclined 
Prisms  if  Bottom  is  Horizontal)  under  Variable  Head. 

CASE  I.  Through  an  orifice  or  short  pipe  in  the  bottom  and 
opening  into  the  air.  —  Fig.  591.     As  the  upper  free  surface, 


738  MECHANICS   OF   ENGINEERING. 

of  area  =  F',  sinks,  F'  remains  constant.  Let  z  =  head  of 
water  at  any  stage  of  the  emptying ;  it  =  20  at  the  outset,  and 
=  0  when  the  vessel  is  empty.  At  any 


<a 


=  I  ?       ^ 
^  1  o  F  ;= 


instant,  Q,  the  rafe  of  discharge  (=  vol- 
ume per  time-unit)  depends  on  s  and  is 

-  ...  (i) 


)j  where  /*  =  coefficient  of  efflux  =  <p  C  = 

FIG.  591.  coefficient  of  velocity  X  coefficient  of  con- 

traction [see  §  495,  eq.  (3)].  "We  here  suppose  7^  so  large 
compared  with  F,  the  area  of  the  orifice,  that  the  free  surface 
of  the  water  in  the  vessel  does  not  acquire  any  notable  velocity 
at  any  stage,  and  that  hence  the  rate  of  efflux  is  the  same  at 
any  instant,  as  for  a  steady  flow  with  head  =  z  and  a  zero 
velocity  in  the  free  surface.  ^  is  considered  constant.  From 
(1)  we  have 


dV=  (vol.  discharged  in  time  dt)  =  Qdt  =  pFVz  dt.  .  (2 

But  this  is  also  equal  to  the  volume  of  the  horizontal  laminar 
F'dz,  through  which  the  free  surface  has  sunk  in  the  same 
time  dt. 

—  F' 

.-.  dt  =     ^  ,—  z~Wz.  .  (3) 


We  have  written  minus  F'dz  because,  dt  being  an  increment, 
dz  is  a  decrement.  To  reduce  the  depth  from  ZQ  (at  the  start, 
time  =  t  =  zero)  to  zn,  demands  a  time 


F'  Czn 

1=--  \ 

pFV*gJ*t 


whence,  by  putting  zn  =  0,  we  have  the  time  necessary  to 
empty  the  whole  prism 


p 
LO 


_     %F'z*  %F'ZS__        2  X  volume  of  vessel  .  (g 

"  ~  Zz  "~  initial  rate  of  discharge  ' 


TIME   OF   EMPTYING   VESSELS. 


730 


that  is,  to  empty  the  vessel  requires  double  the  time  of  dis- 
charging the  same  amount  of  water  if  the  vessel  had  been  kept 
full  (at  constant  head  =  z0  =  altitude  of  prism). 

To  fill  the  same  vessel  through  an  orifice  in  the  bottom,  the 
flow  through  which  is  supplied  from  a 
body  of  water  of  infinite  extent  hori- 
zontally, as  with  the  (single)  canal  lock 
of  Fig.  592,  will  obviously  require  the 
same  time  as  given  in  eq.  (5)  above, 
since  the  effective  head  z  diminishes 
from  ZQ  to  0,  according  to  the  same  law. 

EXAMPLE. — What  time  will  be  needed 
to  empty  a  parallelopipedical  tank  (Fig.  591)  4  ft.  by  5  ft.  in 
horizontal  section  and  6  ft.  deep,  through  a  stop-cock  in  the 
bottom  whose  coefficient  of  efflux  when  fully  open  is  known 
to  be  }Ji  =  0.640,  and  whose  section  of  discharge  is  a  circle  of 
diameter  —  -J  in.  ?  From  given  dimensions  F'  =  4  X  5  =  20 
sq.  ft.,  while  z0  =  6  ft.  Hence  from  eq.  (5)  (ft.-lb.-sec.) 


J£^44s^S^ 


FIG.  592. 


time  of   \ 

emptying  )       0.64  X 


2  X  20  X 


X  32 


3=1  = 


13620  seconds 

g hours  ^min.  Q  sec. 


CASE  II.  Two  communicating  prismatic  vessels.     Required 
the  time  for  the  water  to  come  to  a  common  level  ON,  Yig. 

^ 593,  efflux  taking  place  through  a  small 

orifice,  of  area  =  F,  under  water.     At 
any  instant  the  rate  of  discharge  is 


rf  T 

F 

F      '  r\ 

=  t  1 

"1  !• 

!»  I  i 

rr^ 

IM: 

as  before,    z  =  difference  of  level.   Now 
if  F'  and  7^"  are  the  horizontal  sectional 
areas  of  the  two  prismatic  vessels  (axes 
vertical)  we  have  Ffx  =  F"y,  and  hence  z,  which  =  x  -f-  y> 


FIG.  593. 


a?  = -^r ,     and    <fa?  = 

i+j* 


dz 


740  MECHANICS   OF  ENGINEERING. 

As  before,  we  have 


,    or     dt=—  ; 

' 


—  F'dx  =  — 

F'+F  ' 

Hence,  integrating,  the  time  for  the  difference  of  level  to 
change  from  20  to  zn 


F'+F 
and  by  making  zn  —  0  in  (6),  we  have  the 

C)  Tfl  Tf'f  -\  /~y~ 

time  of  coming  to  a  common  level  =  p,*-™,,  •  ~~p\/  ir~'   CO 


ALGEBEAIC  EXAMPLE. — In  the  double  lock  in  Fig.  594,  let 
L1  be  full,  while  in  L"  the  water  stands  at  a  level  MN  the 

same  as  that  of  the  tail- 
F"  ^  water.  F'  and  F"  are  the 
horizontal  sectional  areas  of 
the  prismatic  locks.  Let 
the  orifice,  0,  between 
them,  be  at  a  depth  =  7i, 
below  the  initial  level  KK 
of  L ',  and  a  height  =  Aa 
above  that,  MN,  of  L" . 
The  orifice  at  0,  area  =  F,  being  opened,  efflux  from  L'  be- 
gins into  the  air,  and  the  level  of  L"  is  gradually  raised  from 
MN  to  OD,  while  that  of  L'  sinks  from  J^S1  to  AB  a  distance 
=  a,  computed  from  the  relation  vol.  F'a  =  vol.  F'!li^,  and 
the  time  occupied  is  [eq.  (4)] 


FIG.  594. 


(8) 


As  soon  as  0  is  submerged,  efflux  takes  place  under  water,  and 
we  have  an  instance  of  Case  II.  Hence  the  time  of  reaching 
a  common  level  (after  submersion  of  0)  (see  eq.  7)  is 


and  the  total  time  is  =  t,  +  t. ,  with  a  = 


TIME   OP    EMPTYING   VESSELS. 


741 


through  a, 


CASE  III.  Emptying  a  vertical  prismatic 
rectangular  "notch"  in  the  side,  or  over- 
fall. —  Fig.  595.  As  before,  let  even  the 
initial  area  (=  zjb)  of  the  notch  be  small 
compared  with  the  horizontal  area  F'  of 
tank.  Let  z  =  depth  of  lower  sill  of  notch 
below  level  of  tank  surface  at  any  instant, 
and  b  =  width  of  notch.  Then,  at  any  in- 
stant (see  eq.  10,  §  504), 


Rate  ofdisch.  (vol.)  =  Q  =  \p 
.-.  vol.  of  disch.  in  dt  = 


and  putting  this  =  —  F'dz  =  vol.  of  water  lost  by  the  tank 
in  time  dt,  we  have 


FIG.  595. 


3     F' 

dt=  —  -  z~ 


whence 


pi 
LO 


3     F,     pl      4 

=  —  ---  —      -  ; 
U  -  1 


i.e., 


Vz 


as  the  time  in  which  the  tank  surface  sinks  from  a  height  5r0 
above  sill  to  a  height  zn  above  sill.  If  we  inquire  the  time  tf 
for  the  water  to  sink  to  the  level  of  the  sill  of  the  notch  we 
put  zn  =  zero,  whence  t'  =  infinity.  As  explanatory  of  this 
result,  note  that  as  z  diminishes  not  only  does  the  velocity  of 
flow  diminish,  but  the  available  area  of  efflux  (=  zb)  also  grows 
less,  whereas  in  Cases  I  and  II  the  orifice  of  efflux  remained 
of  constant  area  =  F. 

Eq.  (10)  is  applicable  to  the  waste-weir  of  a  large  reservoir 
or  pond. 


534.  Time  of  Emptying  Vessels  of  Variable  Horizontal  Sec- 
tions. —  Considering  regular  geometrical  forms  first,  let  us  take 


742 


MECHANICS   OF   ENGINEERING. 


CASE  I.    Wedge-shaped  vessel,  edge  horizontal  and  under- 

neath, orifice  F  in  the  edge,  so  that 
z,  the  variable  head,  is  always  the 
altitude  of  a  triangle  similar  to  the 
section  ABC  'of  the  body  of  water 
when  efflux  begins.  At  any  instant 
during  the  efflux  the  area,  8,  of 
the  free  surface,  variable  here, 
takes  the  place  of  Ff  in  eq.  (3)  of 
§  533,  whence  we  have, 


for  any  case  of  variable  free  surface,  dt  = 


In  the  present  case  S  =  ul,  and  from  similar  triangles 

u  :  z  ::  b  :  z0; 
whence 


(11) 


and 


-  Mz*dz  . 

~=  5 


t  — 


U 


[*.«-«/|,.(18) 


and  hence  the  time  of  emptying  the  whole  wedge,  putting 
zn  =  0,  is 

%blz0       _  4  Vol.  of  wedge 


_  4 


initial  rate  of  discharge 


(13) 


»  jtFVZgz. 

i.e.,  f  as  long  as  to  discharge  the  same  volume  of  water  under 
a  constant  head  =  z0 .     This  is  equally  true  if  the  ends  of  the 
wedge  are  oblique,  so  long  as  they  are  parallel. 
CASE  II.   Right  segment  of  paraboloid  of 
revolution. — Fig.    597.     Axis   vertical.     Ori- 
fice at  vertex.     Here  the  variable  free  surface 
has  at  any  instant  an  area,  =  S,  =  TTU\  u  be- 
ing  the  radius  of    the  circle  and   variable. 
From  a  property  of  the  parabola 


TIME   OF  EMPTYING  VESSELS. 

and  hence,  from  eq.  (11), 

,, 

dt= 


743 


whence,  putting  zn  =  0,  we  have  the  time  of  emptying  the 
whole  vessel 


total  vol. 


,_,  ., 


initial  rateof  disch 


same  result  as  for  the  wedge,  in  Case  I ;  in  fact,  it  applies  to 
any  vessel  in  which  the  areas  of  horizontal  sections  vary 
directly  with  their  heights  above  the  orifice. 

CASE  III.  Any  pyramid  or  cone  •  vertex  down  ;  small  ori- 
fice in  vertex. — Fig.  598.  Let  area  of  the 
base  —  $0 ,  at  upper  edge  of  vessel.  At 
any  stage  of  the  flow  S  =  area  of  base  of 
pyramid  of  water.  From  similar  pyra- 
mids 


So 


FIG.  598. 


and  [eq.  (11)] 


whence  (zn  =  0)  the  time  of  emptying  the  whole  vessel  is 


_  6          Total  volume 
0 


,^  ^ 


5   initial  rate  of  disch* 
CASE  IV.  Sphere.  —  Similarly,  we  may  show  that  to  empty 


744 


MECHANICS   OF   ENGINEERING. 


a  sphere,  of  radius  =  /',  through  a  small  orifice,  of  area 
in  lowest  part,  the  necessary  time  is 


TIT 


15 


8  -Vol. 

5   init.  rate  of  disch. 


535.  Time  of  Emptying  an  Obelisk-shaped  Vessel. — (An  obe- 
lisk may  be  defined  as  a  solid  of  six  plane  faces,  two  of  which 
are  rectangles  in  parallel  planes  and  with  sides  respectively 
parallel,  the  others  trapezoids ;  a  frustum  of  a  pyramid  is  a 
particular  case.) 

A  volume  of  this  shape  is  of  common  occurrence;  see  Fig. 
599.  Let  the  altitude  =  A,  the  two  rectangular  faces  being 
horizontal,  with  dimensions  as  in  figure.  By  drawing  through 

F,  G,  and  II  right  lines  par- 
allel to  EC,  to  cut  the  upper 
base,  we  form  a  rectangle 
KLMC  equal  to  the  lower 
base.  Produce  ML  to  P  and 
KL  to  N,  and  join  PG  and 
NG.  We  have  now  sub- 
divided the  solid  into  a  paral- 
lelepiped KLMC  -  EHGF, 
a  pyramid  PBNL  -  G,  and 
two  wedges,  viz.  APLR-HG  and  LNDM-FG,  with 
their  edges  horizontal ;  and  may  obtain  the  time  necessary  to 
empty  the  whole  obelisk-volume  by  adding  the  times  which 
would  be  necessary  to  empty  the  individual  component  vol- 
umes, separately,  through  the  same  orifice  or  pipe  in  the  bot- 
tom plane  EG.  These  have  been  already  determined  in  the 
preceding  paragraphs.  The  dimensions  of  each  component 
volume  may  be  expressed  in  terms  of  those  of  the  obelisk,  and 
all  have  a  common  altitude  =  h. 

Assuming  the  orifice  to  be  in  the  bottom,  or  else  that  the 
discharging  end  of  the  pipe,  if  such  is  used,  is  in  the  plane  of 
the  bottom  EG^  we  have  as  follows,  F  being  the  area  of  di&- 
charge : 


FIG.  599. 


TIME   OF   EMPTYING   KESERVOIES.  745 

Time  to  empty  the  parallelopiped  }        ^  _     2^       .-=-    . 
separately  would  be  (Case  /,  §  533)  )  *  F  V 


Time  to  empty  the  two  )  07/7      7x17/7,      *  \ 

wedges      separately  K  =  -  •  W  ~  'J  +  ***  ~  *i)  >rr      /ox 

'' 


For  the  pyramid  }  f    _2    (I  —  ^)  @  —  &,)    /7- 

(Case  III,  §  534)  f  '        ^-5* 


Hence  to  empty  the  whole  reservoir  we  have 


.e., 

.     .     (4) 


EXAMPLE.  —  Let  a  reservoir  of  above  form,  and  with  &  =  50  ft., 
I  =  60  ft.,  5j  =  10  ft.,  I,  =  20  ft.,  and  depth  of  water  h  =  16 
ft.,  be  emptied  through  a  straight  iron  pipe,  horizontal,  and 
leaving  the  side  of  the  reservoir  close  to  the  bottom,  at  an  angle 
a  •=.  36°  with  the  inner  plane  of  side.  The  pipe  is  80  ft.  long 
and  4  inches  in  internal  diameter  ;  and  of  clean  surface.  The 
jet  issues  directly  from  this  pipe  into  the  air,  and  hence 
F=\n($f  sq.  feet  To  find  /<,  the  "coefficient  of  efflux" 
(=  0,  the  coefficient  of  velocity  in  this  case,  since  there  is  no 
contraction  at  discharge  orifice),  we  use  eq.  (4)  (the  first  radical) 
of  §  518,  with  f  approx.  =  .006,  and  obtain 


V 


(N.B.    Since  the  velocity  in  the  pipe  diminishes  from  a 
value 

v  =  .361  4/20r  X  16  =  11.6  ft.  per  sec. 

at  the  beginning  of  the  flow  to  v  =  zero  at  the  close,  f  =  .006 
is  a  reasonably  approximate  average  with  which  to  compute 
the  average  0  above  ;  see  §  517. 


746  MECHANICS   OF   ENGINEERING. 

Hence  from  eq.  (4)  of  this  paragraph  (ft.-lb.-sec.  system) 
_  [3  X  50  X  60  +  8  X  10  X  20  +  2(50  X  20+20  X  60)]2 
15  X  0.361  xVax  32.2 


=  31630  sec.  =  8  hrs.  47  min.  10  sec.  i  *ro™>lZ  J1^**  °r 

(        3$  of  the  truth. 

536.  Time  of  Emptying  Reservoirs  of  Irregular  Shape,  Simp- 
son's Rule.  —  From  eq.  (11),  §  534,  we  have,  for  the  time  in 
which  the  free  surface  of  water  in  a  vessel  of  any  shape  what- 
ever sinks  through  a  vertical  distance  =d&9 

dt  =  -  —-^--,  whence   \  time  =  --  —    /  Sz~*dz,  .  .  (1) 
^F^g  '  U=,o       lifVSgJ^ 


where  S  is  the  variable  area  of  the  free  surface  at  any  in- 
stant, and  z  the  head  of  water  at  the  same  instant,  efflux 
proceeding  through  a  small  orifice  (or  extremity  of  pipe)  of 
area  •=.  F.  If.  S  can  be  expressed  in  terms  of  z,  we  can  in- 
tegrate eq.  (1)  (i.e.,  provided  that  Sz~l  has  a  known  anti- 
derivative);  but  if  not,  the  vessel  or  reservoir  being  irregular 
in  form,  as  in  Fig.  600  (which  shows  a  pond  whose  bottom 
has  been  accurately  surveyed,  so  that  we  know  the  value  of  S 
for  any  stage  of  the  emptying),  we  can  still  get  an  approximate 

solution  by  using  Simpson's 
Rule  for  approximate  inte- 
gration. Accordingly,  if  we 
inquire  the  time  in  which 
the  surface  will  sink  from  0 
to  the  entrance  Eoi  the  pipe 
in  Fig.  600  (any  point  n  ;  at 
E.  or  short  of  that),  we 
divide  the  vertical  distance 

from  0  to  n  (4  in  this  figure)  into  an  even  number  of  equal 
parts,  and  from  the  known  form  of  the  pond  compute  the  area 
$  corresponding  to  each  point  of  division,  calling  them  $„,  $,, 
etc.  Then  the  required  time  is  approximately 


TIME   OF   EMPTYING   POND.  747 


*     i        4     r 

+- 


600.  Suppose  we  have  a  pipe  Em  of  the 
same  design  as  in  the  example  of  §  535,  and  an  initial  head  of 
20  =  16  ft.,  so  that  the  same  value  of  /*,  =  .361,  may  be  used. 
Let  zn  —  z0  =  8  feet,  and  divide  this  interval  (of  8  ft.)  into 
four  equal  vertical  spaces  of  2  ft.  each.  If  at  the  respective 
points  of  division  we  find  from  a  previous  survey  that  $Q  = 
400000  sq.  ft.,  S,  =  320000  sq.  ft.,  #,  =  270000  sq.  ft.,  &,  = 
210000  sq.  ft.,  and£4  =  180000  sq.  ft.  ;  while  n  =  4,  p  —  .361, 
and  the  area  F=  i^)2  =  .0873  sq.  ft.,  we  obtain  (ft.,lb.,  sec.) 


G_  _  16-8  ptOOOOO       4  X  320000 

~  0.361  X.  0873  I/2^<~32^X  3X4  L    ylg"  " 


2  X  270000       4  X  210000      ISOOOOn  =  2444000  sec. 
"  ^8"  J=  28d-  6h'  53m' 


volume  discharged,  T7',  may  also  be  found  by  Simpson's 
Rule,  thus  :  Since  each  infinitely  small  horizontal  lamina  has 
a  volume 


or,  approximately, 

4S       $S       IS 


Hence  with  n  =  4  we  have  (ft.,  lb.,  sec.) 


+  180000~|  =  2,160,000  cub.  ft. 


748 


MECHANICS   OF   ENGINEEBING. 


537.  Volume  of  Irregular  Reservoir  Determined  by  Observing 
Progress  of  Emptying.  —  Transforming  eq.  (11),  §  534,  we  have 


But  Sdz  is  the  infinitely  small  volume  d  V  oi  water  lost  by 
the  reservoir  in  the  time  di,  so  that  the  volume  of  the  reser- 
voir between  the  initial  and  final  (0  and  n)  positions  of  the 
horizontal  free  surface  (at  beginning  and  end  of  the  time  tn) 
may  be  written 

(1) 


\= 


This  can  be  integrated  approximately  by  Simpson's  Eule,  if 
the  whole  time  of  emptying,  =  tn,  be  divided  into  an  even 

number      of      equal 
•n 


parts,  and  the  values 

^o  5  si  >  z*  5  etc->  °f  the 
head  of  water  noted 
at  these  equal  inter- 
vals of  time  (not  of 
vertical  height).  The 
corresponding  sur- 
face planes  will  not 
Whence  for  the  particular  case 


FIG. 


be  equidistant,  in  general, 
when  n  =  4  (see  Fig.  601) 


J4*]'    '    '    (2> 


CHAPTEE  VII. 

HYDRODYNAMICS  (Continued)— STEADY  FLOW  OF  WATER  IN 
OPEN  CHANNELS. 

538.  Nomenclature, — Fig.  602.  When  water  flows  in  an 
open  channel,  as  in  rivers,  canals,  mill-races,  water-courses, 
ditches,  etc.,  the  bed 
and  banks  being  rigid, 
the  upper  surface  is 
free  to  conform  in 
shape  to  the  dynamic 
conditions  of  each  case, 
which  therefore  regu- 
late to  that  extent  the 
shape  of  the  cross-sec- 
tion. 

In  the  vertical  trans-  FlG- 602> 

verse  section  A  C  in  figure,  the  line  AC  is  called  the  air-profile 
(usually  to  be  considered  horizontal  and  straight),  while  the 
line  ABC,  or  profile  of  the  bed  and  banks,  is  called  the  ivetted 
perimeter.  It  is  evident  that  the  ratio  of  the  wetted  perimeter 
to  the  whole  perimeter,  though  never  <  -J,  varies  with  the 
form  of  the  transverse  section. 

In  a  longitudinal  section  of  the  stream,  EFGH,  the  angle 
made  by  a  surface  filament  EF  with  the  horizontal  is  called 
the  slope,  and  is  measured  by  the  ratio  s  =  h  :  I,  where  I  is  the 
length  of  a  portion  of  the  filament  and  h  =  the  fall,  or  vertical 
distance  between  the  two  ends  of  that  length.  The  angle  be- 
tween the  horizontal  and  the  line  HG  along  the  bottom  is  not 
necessarily  equal  to  that  of  the  surface,  unless  the  portion  of 
the  stream  forms  a  prism ;  i.e.,  the  slope  of  the  bed  does  not 
necessarily  =  s  =  that  of  surface. 

EXAMPLES. — The  old  Croton  Aqueduct  has  a  slope  of  1.10 
ft.  per  mile;  i.e.,  s  —  .000208.  The  new  aqueduct  (for  New 

749 


750  MECHANICS    OF    ENGINEERING. 

York)  has  a  slope  *  =  .000132,  with  a  larger  transverse  section. 
For  large  sluggish  rivers  s  is  much  smaller. 

539.  Velocity  Measurements.  —  Various  instruments  and 
methods  may  be  employed  for  this  object,  some  of  which  are 
the  following : 

Surface-floats  are  small  balls,  or  pieces  of  wood,  etc.,  so 
colored  and  weighted  as  to  be  readily  seen,  and  still  but  little 
affected  by  the  wind.  These  are  allowed  to  float  with  the  cur- 
rent in  different  parts  of  the  width  of  the  stream,  and  the  sur- 
face velocity  c  in  each  experiment  computed  from  c  =.  I  -=-  #, 
where  I  is  the  distance  described  between  parallel  transverse 
alignments  (or  actual  ropes  where  possible),  whose  distance 
-apart  is  measured  on  the  bank,  and  t  =  the  time  occupied. 

Double-floats.  Two  balls  (or  small  kegs)  of  same  bulk  and 
condition  of  surface,  one  lighter,  the  other  heavier  than  water, 

are  united  by  a  slender  chain,  their 
weights  being  so  adjusted  that  the 
light  ball,  without  projecting  notably 
above  the  surface,  buoys  the  other 
ball  at  any  assigned  depth.  Fig.  603. 

It  is  assumed  that  the  combination 

^^^^w/wm^^m  moves  with  a  velocity  c',  equal  to  the 
FIG.  603.  arithmetic  mean  of  the  surface  veloc- 

ity c0  of  the  stream  and  that,  c,  of  the  water  filaments  at  the 
depth  of  the  lower  ball,  which  latter,  <?,  is  generally  less  than 
<?0 .  That  is,  we  have 

c'=%(cQ  +  e)     a»d    •'•   c  =  2c'  —  c9.  .    .     .     (1) 

Hence,  C0  having  been  previously  obtained,  eq.  (1)  gives  the 
velocity  c  at  any  depth  of  the  lower  ball,  c'  being  observed. 

The  floating  staff  is  a  hollow  cylindrical  rod,  of  adjustable 
length,  weighted  to  float  upright  with  the  top  just  visible.  Its 
observed  velocity  is  assumed  to  be  an  average  of  the  velocities 
of-  all  the  filaments  lying  between  the  ends  of  the  rod. 

Woltmann's  Mill ;  or  Tachometer  ;  or  Current-meter,  Fig. 
604,  consists  of  a  small  wheel  with  inclined  floats  (or  of  a  small 


CURRENT-METERS,    ETC. 


751 


FIG.  604. 


"  screw-propeller"  wheel  S)  held  with  its  plane  ~j  to  the  cur- 
rent, which  causes  it  to  re- 
volve at  a  speed  nearly  pro- 
portional to  the  velocity,  c, 
of  the  water  passing  it. 
By  a  screw-gearing  W  oil 
the  shaft,  connection  is 
made  with  a  counting  ap- 
paratus to  record  the  num- 
ber of  revolutions.  Some- 
times a  vane  B  is  attached,. 
to  compel  the  wheel  to  face 
the  current.  It  is  either 
held  at  the  extremity  of  a  pole  or,  by  being  adjustable  along 
a  vertical  staff  fixed  in  the  bed,  may  be  set  at  any  desired  depth 
below  the  surface.  It  is  usually  so  designed  as  to  be  thrown 
in  and  out  of  gear  by  a  cord  and  spring,  that  the  time  of  mak- 
ing the  indicated  number  of  revolutions  may  be  exactly  noted. 
By  experiments  in  currents  of  known  velocities  a  table  or 
formula  can  be  constructed  by  which  to  interpret  the  indica- 
tions of  any  one  instrument ;  i.e.,  to  find  the  velocity  c  of  the 
current  corresponding  to  an  observed  number  of  revolutions, 
per  minute. 

A  peculiar  form  of  this  instrument  has  been  recently  in- 
vented, called  the  Ritcliie-Haskell  Direction-current  Meter, 
for  which  the  following  is  claimed:  uThis  meter  registers 
electrically  on  dials  in  boat 

•'  <  xV..-.  -.-••.•."•'•,•..',-.• 

from  which  used,  the  direction  ..^ 

and   velocity,   simultaneously,  '£.-- 

of  any  current.    Can  be  used  ,':'-^;~- 

in  river,  harbor,  or  ocean  cur-  fi| 
rents/' 

Pitofs  Tube  consists  in  prin-  ^EEr 

ciple  of  a  vertical  tube  open  =£ 
above,  while  its  lower  end,  also 

open,  is  bent  horizontally  up-  ~ 
stream;  see  A  in  figure.   After 

the  oscillations   have   ceased,  the  water  in  the  tube  remains 


FIG.  605. 


752  '  MECHANICS   OF   ENGINEERING. 

stationary  with  its  free  surface  a  height,  A',  above  that  of  the 
stream,  on  account  of  the  continuous  impact  of  the  current 
against  the  lower  end  of  the  column.  By  the  addition  of 
another  vertical  tube  (see  B  in  figure)  with  the  face  of  its 
lower  (open)  end  parallel  to  the  current  (so  that  the  water- 
level  in  it  is  the  same  as  that  of  the  current),  both  tubes  being 
provided  with  stop-cocks,  we  may,  after  closing  the  stop- 
cocks, lift  the  apparatus  into  a  boat  and  read  off  the  height  Ji' 
at  leisure.  We  may  also  cause  both  columns  of  water  to 
mount,  through  flexible  tubes,  into  convenient  tubes  in  the 
boat  by  putting  the  upper  ends  of  both  tubes  in  communica- 
tion with  a  receiver  of  rarefied  air,  and  thus  watch  the  oscilla- 
tions and  obtain  a  more  accurate  value  of  h'.  [See  Van  Nos- 
trand's  Mag.  for  Mar.  '78,  p.  255.]  Theoretically  (see  §  565), 
the  thickness  of  the  walls  of  the  tube  at  the  lower  end  being 
considerable,  we  have 

c  =  VgK (1) 

as  a  relation  between  c,  the  velocity  of  the  particles  impinging 
on  the  lower  end,  and  the  static  height  h'  (§  565).  Eq.  (1)  is 
verified  fairly  well  by  Weisbach's  experiments  with  fine  in- 
struments, used  with  velocities  of  from  0.32  to  1.24:  meters 
per  second.  Weisbach  found 

c  =    3.54  Vhf  (In  meters)   met.  per  sec., 
whereas  eq.  (1)  gives 

c=     3.133  Vh'  (in  meters)    met.  per  sec. 

In  the  instruments  used  by  Weisbach  the  end  of  the  tube 
turning  up-stream  was  probably  straight ;  i.e.,  neither  flaring 
nor  conically  convergent.  A  change  in  this  respect  alters  the 
relation  between  c  and  h ';  see  §  565  for  Pitot's  and  Darcy's 
results. 

Pitot's  Tube,  though  simple,  is  not  so  accurate  as  the  ta- 
chometer. 


CURRENT-METERS.  753 

The  Hydrometric  Pendulum,  a  rather  uncertain  instrument, 
is  readily  understood  from  Fig.  606.  The  side  AB,  of  the 
quadrant  ABC,  being  held  vertical,  the 
plane  of  the  quadrant  is  made  parallel  to 
the  current.  The  angle  6  between  the 
cord  and  the  vertical  depends  on  G,  the 
effective  weight  (i.e.,  actual  weight  dimin- 
ished by  the  buoyant  effort)  of  the  ball 
(heavier  than  water),  and  the  amount  of  jP, 
the  impulse  or  horizontal  pressure  of  the 
current  against  the  latter,  since  the  cord 
will  take  the  direction  of  the  resultant  72,  FlG>  606< 

for  equilibrium. 

Now  P  (see  §  572)  for  a  ball  of  given  size  and  character  of 
surface  varies  (nearly)  as  the  square  of  the  velocity ;  i.e.,  if  Pf 
is  the  impulse  on  a  given  stationary  ball,  when  the  velocity  of 
the  current  ==  c1 ',  then  for  any  other  velocity  c  we  have 


P  —  impulse  —  —  c* (2) 

c 


P 

From  this  and  the  relation  tan  0  =  —^  we  derive 

Cr 


^•^tantf (3) 


With  a  given  instrument  and  a  specified  system  of  units,  the 
numerical  value  of  the  first  radical  may  be  determined  as  a 
single  quantity,  by  experimenting  with  a  known  velocity  and 
the  value  of  6  then  indicated,  and  may  then,  as  a  constant  fac- 
tor, be  employed  in  (3)  for  finding  the  value  of  c  for  any  ob- 
served value  of  0 ;  but  the  same  units  must  be  used  as  before. 

540.  Velocities  in  Different  Parts  of  a  Transverse  Section. — 
The  results  of  velocity-measurements  made  by  many  experi- 
menters do  not  agree  in  supporting  any  very  definite  relation 
between  the  greatest  surface  velocity  (c0  max.)  of  a  transverse 


754  MECHANICS    OF   ENGINEERING. 

section  and  the  velocities  at  other  points  of  the  section,  but 
establish  a  few  general  propositions  : 

1st.  In  any  vertical  line  the  velocity  is  a  maximum  quite 
near  the  surface,  and  diminishes  from  that  point  both  toward 
the  bottom  and  toward  the  surface. 

2d.  In  any  transverse  horizontal  line  the  velocity  is  a  maxi- 
mum near  the  middle  of  the  stream,  diminishing  toward  the 
banks. 

3d.  The  mean  velocity  —  v,  of  the  whole  transverse  section, 
i.e.,  the  velocity  which  must  be  multiplied  by  the  area,  F,  of 
the  section,  to  obtain  the  volume  delivered  per  unit  of  time, 


Q  =  Fv,  .    .    .    .'   .~r.  ,-    (1) 

is  about  83  per  cent  of  the  maximum  surface  velocity  (c0  max.) 
observed  when  the  air  is  still  ;  i.e., 


v  =  0.83  X  (0. 


Of  eight  experimenters  cited  by  Prof.  Bowser,  only  one  gives 
a  value  (=  0.62)  differing  more  than  .05  from  .83,  while  others 
obtained  the  values  .82,  .78,  .82,  .80,  .82,  .83. 

In  the  survey  of  the  Mississippi  River  by  Humphreys  and 
Abbot,  1861,  it  was  found  that  the  law  of  variation  of  the 
velocity  in  any  given  vertical  line  could  be  fairly  well  repre- 
sented by  the  ordinates  of  a  parabola  (Fig.  607)  with  its  axis 
.*••'.: ".••^.:\ :/>. v •.;'•;:• ::.:*,  horizontal  and  its  vertex  at  a  distance  dl 
~'i  -=V- -t^'-"— -i==  below  the  surface  according  to  the  follow- 

X. I  i    _\      , 

'  £  I  -I  I  ,^~^\~    E  ing  relation,  f  being  a  number  dependent 

•^-    .  —   T  —    —    '  ,. i  1  #•  f"        j  1  •         T        /  (*  r\.        (* 


_i_Y  ~  >\cdr-~  on  the  force  of  the  wind  (from  0  for  no 
'  f=L  wind  to  10  for  a  hurricane) : 

=  [0.31T±0.06/"]rf;   .     .     (3) 

=-  where  d  is  the  total  depth,  and  the  double 
sign  is  to  be  taken  -j-  for  an  up-stream,  — 
FIG.  607.  for  a   down-stream,  wind.     The   following 

relations  were  also  based  on  the  results  of  the  survey : 


(putting,  for  brevity,  B  =  1.69  -r-  Yd  +  1.5,)     .     (4) 


VELOCITIES   IN   OPEtf    CHANNEL.  755 

/- .7x2 

-vm?-=^\, (5) 


and 

c^  =  cm  +  -fy^£v.       .-.••.:*'   .    .    (7) 

(These  equations  are  not  of  homogeneous  form,  but  call  for 
the  foot  and  second  as  units.) 
In  (4),  (5),  and  (6), 

c  =  velocity  at  any  depth  &  below  the  surface  ; 
cm  =  mean  velocity  in  the  vertical  curve ; 
cdl  =  max.         "  «  "          " 

Cfa  =  "         at  mid-depth ; 

cd  =  velocity  at  bottom  ; 
v  =  mean  velocity  of  the  whole  transverse  section. 

It  was  also  found  that  the  parameter  of  the  parabola  varied 
inversely  as  the  square  root  of  the  mean  velocity  cm  of  curve. 

In  general  the  bottom  velocity  (<?d)  is  somewhat  more  than  \ 
the  maximum  velocity  (cdl)  in  the  same  vertical.  In  the  Mis- 
sissippi the  velocity  at  mid-depth  in  any  vertical  was  found  to 
be  very  nearly  .96  of  the  surface  velocity  in  the  same  vertical ; 
this  fact  is  important,  as  it  simplifies  the  approximate  gauging 
of  a  stream. 

541.  Gauging  a  Stream  or  River. — Where  the  relation  (eq.  (2), 
§  540)  v  —  .83  (<?0  max.)  is  not  considered  accurate  enough  for 
substitution  in  Q  —  Fv  to  obtain  the  volume  of  discharge  (or 
delivery)  Q  of  a  stream  per  time-unit,  the  transverse  section 
may  be  divided  into  a  number  of  subdivisions  as  in  Fig.  608, 
of  widths  a, ,  «,,  etc.,  and 
mean  depths  dl9  <7a,  etc., 
and  the  respective  mean 
velocities,  c1 ,  <?2 ,  etc.,  com- 
puted from  measurements 
with  current-meters  ;  whence  we  may  write 

Q  =  a,  d,c,  +  atdtct  +  a,d,cz  +  etc.     ...     (7) 


756 


MECHANICS   OF   ENGINEEEING. 


With  a  small  stream  or  ditch,  however,  we  may  erect  a  ver- 
tical boarding,  and  allow 
the  water  to  flow  through  a 
rectangular  notch  or  over- 
fall, Fig.  609,  and  after  the 
head  surface 
permanent, 


has  become 
measure  A2 
(depth  *  of  sill  below  the 
level  surface  somewhat 
back  of  boards),  and  b 
(width)  and  use  the  formu- 
lae of  §  504;  see  examples 


FIG.  609. 


in  that  article. 


542.  Uniform  Motion  in  an  Open  Channel. — We  shall  now 
consider  a  straight  stream  of  indefinite  length  in  which  the 
flow  is  steady,  i.e.,  a  state  of  permanency  exists,  as  distin- 
guished from  a  freshet  or  a  wave.  That  is,  the  flow  is  steady 
when  the  water  assumes  flxed  values  of  mean  velocity  v,  and 
sectional  area  F,  on  passing  a  given  point  of  the  bed  or  bank ; 
and  the 

Eq.  of  continuity  .  .  Q  =  Fv  =  JFJ0v0  —  Flvl  =  constant .  .  (1) 

holds  good  whether  those  sections  are  equal  or  not. 

By  uniform  motion  is  meant  that  (the  section  of  the  bed 
and  banks  being  of  constant  size  and  shape)  the  slope  of  the 
bed,  the  quantity  of  water  (volume  =  Q)  flowing  per  time- 
unit,  and  the  extent  of  the  wetted  perimeter,  are  so  adjusted 
to  each  other  that  the  mean  velocity  of  flow  is  the  same  in  all 
transverse  sections,  and  consequently  the  area  and  shape  of  the 
transverse  section  is  the  same  at  all  points ;  and  the  slope  of 
the  surface  —  that  of  the  bed.  We  may  therefore  consider, 
for  simplicity,  that  we  have  to  deal  with  a  prism  of  water  of 
indefinite  length  sliding  down  an  inclined  rough  bed  of  con- 
stant slope  and  moving  with  uniform,  velocity  (viz.,  the  mean 
velocity  v  common  to  all  the  sections) ;  that  is,  there  is  no  ac- 
celeration. Let  Fig.  610  show,  free,  a  portion  of  this  prism, 
of  length  =  Z,  and  having  its  bases  1  to  the  bed  and  surface. 


UNIFORM   MOTION.       OPBN   CHANNEL.  757 

The  hydrostatic  pressures  at  the  two  ends  balance  each  other 
from  the  identity  of  conditions.     The  only  other  forces  having 


jU-4-JL 


FIG.  610. 


components  parallel  to  the  bed  and  surface  are  the  weight 
G  =  Fly  of  the  prism  (where  y  =  heaviness  of  water)  making 
an  angle  =  s  (  =  slope)  with  a  normal  to  the  surface,  and  the 
friction  between  the  water  and  the  bed  which  is  parallel  to  the 
surface.  The  amount  of  this  friction  for  the  prism  in  question 
may  be  expressed  as  in  §  510,  viz.: 


=fwly^9 


in  which  S  =  wl  =  rubbing  surface  (area)  =  wetted  perimeter, 
w,  X  length  (see  §  538),  and  /"an  abstract  number.  Since  the 
mass  of  water  in  Fig.  610  is  supposed  to  be  in  relative  equili- 
brium, we  may  apply  to  it  the  laws  of  motion  of  a  rigid  body, 
and  since  the  motion  is  a  uniform  translation  (§  109)  the  com- 
ponents, parallel  to  the  surface,  of  all  the  forces  must  balance. 

h  tf 

.'.  G  sin  s  must  =  P  =  fric.  ;     /.   Fly  —  =fwly  —  —  ; 

i  2g 

whence 


or 


in  which  F  '-t-w  is  called  7?,  the  hydraulic  mean  depth,  or 
hydraulic  radius.     (3)  is  sometimes  expressed  by  saying  that 


758 


MECHANICS   OF   ENGINEEEING, 


the  whole  fall,  or  head,  ^,  is  (in  uniform  motion)  absorbed  in 
friction-head.     Also,  since  the  slope  s  =  h  -f- 1,  we  have 


v  =  J'*LV£s;    or,   «  = 


(*> 


which  is  of  the  same  form  as  Chezy's  formula  in  §  519  for  a 
very  long  straight  pipe  (the  slope  s  of  the  actual  surface  in  this 
case  corresponding  to  the  slope  along  piezometer-summits  in 
that  of  a  closed  pipe).  In  (4)  the  coefficient  A  =  V%g  -s-f  is 
not,  like/",  an  abstract  number,  but  its  numerical  value  depends 
on  the  system  of  units  employed. 

542a.  Experiments  on  the  Flow  of  Water  in  Open  Channels. — 
Those  of  Darcy  and  Bazin,  begun  in  1855  and  published  in 
1865  ("  Recherches  Hydrauliques"),  were  very  carefully  con- 
ducted with  open  conduits  of  a  variety  of  shapes,  sizes,  slopes, 
and  character  of  surface.  In  most  of  these  a  uniform  flow  was 
secured  before  the  taking  of  measurements.  The  velocities 
ranged  between  from  about  0.5  to  8  or  10  ft.  per  second,  the 
hydraulic  radii  from  0.03  to  3.0  ft.,  with  deliveries  as  high  as 
182  cub.  ft.  per  second.  For  example,  the  following  results 
were  obtained  in  the  canals  of  Marseilles  and  Craponne,  the 
quantity  A  being  for  the  foot  and  second.  The  sections  were 
nearly  all  rectangular.  See  eq.  (4)  above. 


No. 

(cub.'  ft. 
per  sec.) 

R. 

(ft.) 

s. 
abs. 
numb. 

V. 

(ft.  per 
sec.) 

A. 
(foot  and 
sec.) 

Character  of  the  masonry 
surface. 

1 

182.73 

1.504 

.0037 

10.26 

137.1 

Very  smooth. 

2 

143.74 

1.774 

.00084 

5.55 

125. 

Quite 

3 

43.93 

.708 

.029 

11.23 

78.4 

4 

43.93 

.615 

.060 

13.93 

72.5 

Hammered  stone. 

5 

43.93 

.881 

.0121 

7.58 

73.5 

Rather  rough. 

6 

43.93 

.835 

.014 

8.36 

77.3 

7 

167.68 

2.871 

.  00043 

2.54 

72.2 

Mud  and  vegetation. 

[In  Experiment  No.  7  the  flow  had  not  fully  reached  a  state 
of  permanency.] 

Fteley  and  Stearns's  experiments  on  the  Sudbury  conduit  at 
Boston,  Mass.  (Trans.  A.  8.  C.  E.,  '83),  from  1878  to  1880,  are 
also  valuable.  This  open  channel  was  of  brick  masonry  with 


UNIFORM   MOTION. 

good  mortar  joints,  and  about  9  ft.  wide ;  the  depths  of  water 
ranging  from  1.5  to  4.5  ft.  With  plaster  of  pure  cement  on 
the  bed  in  one  of  the  experiments  the  high  value  of  A  =  153.6 
was  reached  (foot  and  second),  with  v  =  2.805  ft.  per  second, 
R  =  2.111  ft.,  s  =  .0001580,  and  Q  =  87.17  cu.  ft.  per  second. 

Captain  Cunningham,  in  his  experiments  on  the  Ganges 
Canal  at  Roorkee,  India,  in  1881,  found  A  to  range  from  48 
to  130  (foot  and  second). 

Humphreys  and  Abbot's  experiments  on  the  Mississippi 
River  and  branches  (see  §  540),  with  values  of  R  =  from  2  or 
3  ft.  to  72  ft.,  furnish  values  of  A  =  from  53  to  167  (foot  and 
second). 

542b.  Kutter's  Formula. — The  experiments  upon  which 
Weisbach  based  his  deductions  for/*,  the  coefficient  of  fluid 
friction,  were  scanty  and  on  too  small  a  scale  to  warrant  gener- 
al conclusions.  That  author  considered  that/*  depended  only 
on  the  velocity,  disregarding  altogether  the  degree  of  rough- 
ness of  the  bed,  and  gave  a  table  of  values  in  accordance  with 
that  view,  these  values  ranging  from  .0075  for  15  ft.  per  sec. 
to  .0109  for  0.4  ft.  per  sec. ;  but  in  1869  Messrs.  Kutter  and 
Ganguillet,  having  a  much  wider  range  of  experimental  data 
at  command,  including  those  of  Darcy  and  Bazin,  and  those 
obtained  on  the  Mississippi  River,  evolved  a  formula,  known 
as  Gutter's  Formula,  for  the  uniform  motion  of  water  in  open 
channels,  which  is  claimed  to  harmonize  in  a  fairly  satisfactory 
manner  the  chief  results  of  the  best  experiments  in  that  direc- 
tion. They  make  the  coefficient  A  in  eq.  (4)  (or  rather  the 

factor  — —  contained  in  A)  a  function  of  JR,  $,  and  also  n  an 

J 
abstract  number,  or  coefficient  of  roughness,  depending  on  the 

nature  of  the  surface  of  the  bed  and  banks  ;  viz., 


*un 

41.6 

1.811 

.00281 

1       N, 

r 

0 

per  f 
sec.  J 

1+^41.6 

.00281 

J 

n 

1 

VI 

?(in 

feet) 

which  is  Kutter^ s  Formula. 


760  MECHANICS   OF   ENGINEERING. 

That  is,  comparing  (5)  with  (4),  we  have /a  function  of 
It,  and  s,  as  follows : 


^» 


n 


.      .     .     (6) 


From  (6)  it  appears  that  f  decreases  with  an  increasing  j??, 
as  has  been  also  noted  in  the  case  of  closed  pipes  (§  -517) ;  that 
it  increases  with  increasing  roughness  of  surface ;  and  that  it 
is  somewhat  dependent  on  the  slope.  The  makers  of  the 
formula  give  the  following  values  for  n. 

Values  of  n.    n  = 

.009  for  well-planed  timber  bed  ; 

.010  for  plaster  in  pure  cement ; 

.011  for  plaster  in  cement  with  \  sand  ; 

.012  for  unplaned  timber ; 

.013  for  ashlar  and  brickwork ;  * 

.015  for  canvas  lining  on  frames ; 

.017  for  rubble ; 

.020  for  canals  in  very  firm  gravel ; 

.025  for  rivers  and  canals  in  perfect  order  and  regimen,  and 
perfectly  free  from  stones  and  weeds ; 

.030  for  rivers  and  canals  in  moderately  good  order  and  regi- 
men, having  stones  and  weeds  occasionally ; 

.035  for  rivers  and  canals  in  bad  order  and  regimen,  overgrown 
with,  vegetation  and  strewn  with  stones  or  detritus  of 
any  sort. 

Kutter's  Formula  is  claimed  to  apply  to  all  kinds  and  sizes  of 
watercourses,  from  large  rivers  to  sewers  and  ditches  ;  for  uni- 
form motion.  If  4/7?  is  the  unknown  quantity,  Kutter's  For- 
mula leads  to  a  quadratic  equation  ;  if  s  the  slope,  to  a  cubic. 
Hence,  to  save  computation,  tables  have  been  prepared,  some 
of  which  will  be  found  in  vol.  28  of  Yan  Nostrand's  Magazine 

*  For  ordinary  brick  sewers  Mr.  R.  F.  Hartford  claims  that  n  =  .014 
gives  good  results.  See  Jour.  Eng.  Societies  for  '84- '85,  p.  220. 


UNIFORM    MOTION.      OPEN   CHANNEL. 


761 


(pp.  135  and  393)  (sewers),  and  in  Jackson's  works  on  Hydrau- 
lics (rivers). 

The  following  table  will  give  the  student  an  idea  of  the 


variation   of   the   coefficient  A,  = 


-£-,  of  eq.  (4),  or  large 

t/ 


bracket  of  eq.  (5),  with  different  hydraulic  radii,  slopes,  and 
values  of  ?&,  according  to  Kutter's  Formula ;  from  R  =  £  ft., 
for  a  small  ditch  or  sluice-way  (or  a  wide  and  shallow  stream), 
to  R  —  15  ft.,  for  a  river  or  canal  of  considerable  size.  Under 
each  value  of  R  are  given  two  values  of  A  ;  one  for  a  slope  of 
s'  —  .001,  and  the  other  for  s"  =  .00005.  All  these  values  of 
A  imply  the  use  of  the  foot  and  second. 

These  values  of  A  have  been  scaled  by  the  writer  from  a 
diagram  given  in  Jackson's  translation  of  Kutter's  "  Hydraulic 
Tables^  and  are  therefore  only  approximate.  The  corre- 
sponding values  of  /*,  the  coefficient  of  fluid  friction,  can  be 

computed  from  f  =  — ^ . 


n 

0.010 
0.015 
0.020 
0.025 
0.030 
0.035 

R  =  *  f  t. 

R  =  1  f  t. 

R  =  3  f  t. 

R  =  6  f  t. 

R  =  15  ft. 

for  for 
s'  s" 

for           for 
s'             s" 

for           for 
s'             s" 

for           f  el- 
s'            s" 

for      for 
s'        s" 

133  104 
83  68 
58  49 
45  38 
36  31 
28  25 

149        137 

96          87 
70          63 
54          48 
43          40 
38          34 

174        174 
118        118 
87          87 
70          70 
58          58 
50          50 

187        196 
128        137 
98        106 
80          85 
66          72 
58          64 

199    222 
138    158 
110    126 
90    106 
77      90 
68      81 

The  formula  used  in  designing  the  New  Aqueduct  for  New 
York  City,  in  1885,  by  Mr.  Fteley,  consulting  engineer,  was 

[see  (4)]  "  ' 

v  (ft.  per  sec.}  =  142  VE(irift.)  X  s,     ...     (7) 

whereas  Kutter's  Formula  gives  for  the  same  case  (a  circular 
section  of  14  ft.  diameter,  and  slope  of  0.7  ft.  to  the  mile), 
with  n  =  0.013, 


v(ft  per  xec.)  =  140.7  V E(ln  ft.}  X  *.  .    .     .     (8) 


762  MECHANICS   OF   ENGINEERING. 

To  quote  from  a  letter  of  Mr.  I.  A.  Shaler  of  the  Aqueduct 
Corps  of  Engineers,  "  Mr.  Fteley  states  that  the  cleanliness  of 
the  conduit  (Sudbury)  had  much  to  do  in  affecting  the  flow. 
He  found  the  flow  to  be  increased  by  7  or  8  per  cent  in  a  por- 
tion which  had  been  washed  with  a  thin  wash  of  Portland 
cement." 

EXAMPLE  1. — A  canal  1000  ft.  long  of  the  trapezoidal  sec- 
tion in  Fig.  611  is  required  to  deliver  300  cubic  ft.  of  water 
per  second  with  the  water  8  ft.  deep  at  all 
sections  (i.e.,  with  uniform  motion),  the 
slope  of  the  bank  being  such  that  for  a  depth 
of  8  ft.  the  width  of  the  water  surface  (or 
length  of  air-profile)  will  be  20  ft.;  and  the 
coefficient  for  roughness  being  n  —  .020.  What  is  the  neces- 
sary slope  to  be  given  to  the  bed  (slope  of  bed  =  that  of  sur- 
face, here)  (ft.,  lb.,  sec.)  \ 
The  mean  velocity 

•v  =  Q  +  F=  300  -f-  i  (20  +  8)  8  =  2.6T  ft.  per  sec. 

[So  that  the  surface  velocity  of  mid-channel  in  any  section 
would  probably  be  (<?omax.)  —  v  -r-  0.83  =  3.21  ft.  per  sec.  (eq. 
(2),  §  540).] 

The  wetted  perimeter 


w  =  8  +  2  1/8'  +  62  =  28  ft., 
and  therefore  the  mean  hydraulic  depth 

=  E  =  F+  w  =  112  -f-  28  =  4  ft. 

To  obtain  a  first  approximation  for  the  slope,  we  may  use 
the  value/'  =  .00795  given  by  Weisbach  for  a  velocity  of  2.67 
ft.  per  sec.,  and  obtain,  from  (3), 

__  .00795  X  1000  X  28  (2.67)*  _  ,    , 

112  X  2  X  32.2 

i.e.,  s  =  7i-irl  =  .000221. 


UNIFORM   MOTION   IN    OPEN   CHANNEL. 


763 


With  this  value  for  the  slope  and  7? .=  4  ft.  (see  above),  we 
then  have,  from  eq.  (6)  (putting  n  =  .020), 


.00281  \   .020 


.020        .000221 


=  .0071, 


with  which  value  of  f  we  now  obtain 

h  =  0.200  feet ;     i.e.,    slope  =  s  =  .00020. 

EXAMPLE  2. — If  the  bed  of  a  creek  falls  20  inches  every 
1500  ft.  of  length,  what  volume  of  water  must  be  flowing  to 
maintain  a  uniform  mean  depth  of  4^  ft.,  the  corresponding 
surface-width  being  40  ft.,  and  wetted  perimeter  46  ft.  ?  The 
bed  is  "  in  moderately  good  order  and  regimen  ;"  use  Kutter's 
Formula,  putting  n  =  0.030  (ft.  and  sec.). 

First  we  have 


X 


-^-(46  X 


while  VIZ  (ft.)  =1.98,  and  the  slope  =  s  =  f-f  -j- 1500=.00111 ; 
hence 


or 


104.43  X  .066 
1.6685 


Hence,  also, 


1   .OOlllJ  1.98 
v  =  4.13  ft.  per  sec. 


=  ^3.4  cub.  ft.  per  sec. 


[N.B.  Weisbach  works  this  same  example  by  eq.  (3)  with  a 
value  of  /taken  from  his  own  table,  his  result  being  v  =  6.1 


764  MECHANICS    OP   ENGINEERING. 

ft.  per  sec.,  which  would  probably  be  attained  in  practice  only 
by  making  the  bed  and  banks  smoother  than  as  given.] 
EXAMPLE  3. — The  desired  transverse  water-section  of  a  canal 
is  given  in  Fig.  612.  The  slope  is  to  be 
3  ft.  in  1600 ;  i.e.,  s  —  3  -h  1600 ;  or,  for 
I  —  1600  ft.,  h  —  3  ft.  What  must  be  the 
velocity  (mean)  of  each  section,  for  a  uni- 
form motion  •  the  corresponding  volume 
delivered  per  sec.,  Q,  —  Fv,  =  ? ;  assuming  that  the  character 
of  the  surface  warrants  the  value  n  —  .030  2 

Knowing  the  slope  s9  =  3  ~  1600  ;  and  the  hydraulic  radius 
R,  =  F-+-  w,  =  79.28  sq.  ft.  H-  24.67  ft.,  =  3.215  feet ;  with 
n  =  .030  we  substitute  directly  in  eq.  (5),  obtaining  v  —  4.67 
ft.  per  sec. ;  whence  Q  =  Fv  —  370  cub.  ft.  per  sec. 

543.  Hydraulic  Mean  Depth  for  a  Minimum  Frictional  Resist- 
ance.— We  note,  from  eq.  (3),  §  542,  that  if  an  open  channel 
of  given  length  I  and  sectional  area  F  is  to  deliver  a  given 
volume,  Q,  per  time-unit  with  uniform  motion,  so  that  the 
common  mean  velocity  v  of  all  sections  (=  Q  -+-  F)  is  also  a 
given  quantity,  the  necessary  fall  —  A,  or  slope  s  —  h  -~-  I,  is 
seen  to  be  inversely  proportional  to  ./?,  the  hydraulic  mean 
depth  of  the  section,  =  (F  -±-  w\  —  sectional  area  -f-  wetted 
perimeter. 

For  h  to  be  as  small  as  possible,  we  may  design  the  form  of 
transverse  section,  so  as  to  make  R  as  large  as  possible ;  i.e., 
to  make  the  wetted  perimeter  a  minimum  for  a  given  F',  for 
in  this  way  a  minimum  of  frictional  contact,  or  area  of  rub- 
bing surface,  is  obtained  for  a  prism  of  water  of  given  sectional 
area  J^and  given  length  I. 

In  a  closed  pipe  running  full  the  wetted  perimeter  is  the 
whole  perimeter ;  and  if  the  given  sectional  area  is  shaped  in 
the  form  of  a  circle,  the  wetted  perimeter,  =  10,  is  a  minimum 
(and  R  a  maximum).  If  the  full  pipe  must  have  a  polygonal 
shape  of  n  sides,  then  the  regular  polygon  of  n  sides  will  pro- 
vide a  minimum  w. 

Whence  it  follows  that  if  the  pipe  or  channel  is  running 


UNIFORM   MOTION   IN   OPEN   CHANNEL. 


765 


half  full,  and  thus  becomes  an  open  channel,  the  semicircle, 

of  all  curvilinear  water  pro- 
files, gives  a  minimum  w. 
Also,  of  all  trapezoidal  pro- 
files with  banks  at  60°  with 
the  horizontal  the  half  of  a 
regular  hexagon  gives  a 
minimum  w.  Among  all 
rectangular  sections  the  half 

FIG  613.  •  •     • 

square  gives  a  minimum  w  ; 

and  of  all  half  octagons  the  half  of  a  regular  octagon  gives  a 
minimum  w  (and  max.  H)  for  a  given  F.  See  Fig.  613  for 
all  these. 

The  egg-shaped  outline,  Fig.  614,  small  end  down,  is  fre- 
quently given  to  sewers  in  which  it  is  important  that  the 
different  velocities  of  the  water  at  dif- 
ferent stages  (depths)  of  flow  (depend- 
ing on  the  volume  of  liquid  passing  per 
unit-time)  should  not  vary  widely  from 
each  other.  The  lower  portion  ABC, 
providing  for  the  lowest  stage  of  flow 
AB,  is  nearly  semicircular,  and  thus  in- 
duces a  velocity  of  flow  (the  slope  being 
constant  at  all  stages)  which  does  not 
differ  extremely  from  that  occurring 
when  the  water  flows  at  its  highest 
stage  DE,  although  this  latter  velocity  is  the  greater;  the 
reason  being  that  ABC  from  its  advantageous  form  has  a 
hydraulic  radius,  R,  larger  in  proportion  to  its  sectional  area, 
F,  than  DCE. 

That  is,  F  -r-  w  for  ABC  is  more  nearly  equal  to  F  -^  w  for 
DEC  than  if  DEC  were  a  semicircle,  and  the  velocity  at  the 
lowest  stage  may  still  be  sufficiently  great  to  prevent  the  de- 
posit of  sediment.  See  §  575. 

544.  Trapezoid  of  Fixed  Side-slope. — For  large  artificial  water- 
courses and  canals  the  trapezoid,  or  three-sided  water-profile 
(symmetrical),  is  customary,  and  the  inclination  of  the  bank, 


FIG.  614. 


766  MECHANICS    OF   ENGINEERING. 

or  angle  6  with  the  horizontal,  Fig.  615,  is  often  determined 

by  the  nature    of    the   material 
composing  it,   to  guard  against 
washouts,    caving   in,   etc.     We 
are  therefore  concerned  with  the 
mKZttZZiSSZi'f       following  problem :     Given  the 
FlG- 615-  area,  F,  of  the  transverse  section, 

and  the  angle  0,  required  the  value  of  the  depth  x  (or  of  upper 
width  z,  or  of  lower  width  y,  both  of  which  are  functions  of  x) 
to  make  the  hydraulic  mean  depth,  R  —  F  -f-  w,  a  maximum, 
or  w  -r-  F  a  minimum...     F  is  constant. 
From  the  figure  we  have 


w  =  AB  +  1BC  =  y  +  Zx cosec.  6,   .     .     .     (1) 
and 

F  =  yx  4-  a?2  cot.  0 ; 
whence 

y  =  -.(F—  x*  cot.  0), (2) 

x 

substituting  which   in    (1)  and  dividing   by   F,   noting  that 
2  cosec.  9  —  cot.  0  = 


sin  B 

1        1 

,    2  —  cos  6 

J- 

x                   • 

jR       x 

7<7sin  0 

IV  JL  J-  ,         fJ     VvV/0     I/  /O\ 

jr  =  -77  =  ~  +  —v^^r  'x (3) 


For  a  minimum  w  we  put 


dx 


1        2  —  cos  0 
=  0;     i.e.,     --T+     y0:ri/a"  =  °; 


/.    a?  (for.  max.  or  min.  w)  =  ±  4/5 —      —ft 

The  -f-  sign  makes  the  second  derivative  positive,  and  hence 
for  a  min.  w  or  max.  R  we  have 

x  (call  it  a/)  =  x'  =  • — :          =r-,     .     .     .     (4) 
V2  —  cos  0 


TRAPEZOID   FOR   MINIMUM    WETTED   PERIMETER.     767 

while  the  corresponding  values  for  the  other  dimensions  are 

y'  =  -^-—  oj'  cot.  #   .........     (5) 

ilC 

and 

s'=y'  +  2x'cot.  &=—  +  %'  cot.  0.    .     .     .     (6) 

vu 

For  the  corresponding  hydraulic  mean  depth  R'  [see  (3)], 
i.e.,  the  max.  R,  we  have 


1     .2  —  0080    ,_    2 
' 


x'         Fsm  0  x" 

f*™°o (8> 


Equations  (4),  (5),  ...  (8)  hold  good,  then,  for  the  trapezoi- 
dal section  of  least  f  rictional  resistance  for  a  given  angle  6. 

PKOBLEM.  —  Required  the  dimensions  of  the  trapezoidal  sec- 
tion of  minimum  f  rictional  resistance  for  6  =  45°,  which  with 
k  =  6  inches  fall  in  every  1200  feet  (—  Z)  is  required  to  de- 
liver Q  =  360  cub.  ft.  of  water  per  minute  with  uniform 
motion. 

Here  we  have  given,  with  uniform  motion,  A,  Z,  and  Qr 
with  the  requirement  that  the  section  shall  be  trapezoidal,  with 
6  =  45°,  and  of  minimum  frictional  resistance.  The  following 
equations  are  available  : 

Eq.  of  continuity  .  .  .  Q  =  Fv,  ......     (1;) 


Eq.  (8)  preceding,  for  con-  )       7?'—  A      /     s^ 
dition  of  least  resistance   j  2    y   2  — 


cos  6 


.  (2'Y 


There  are  three  unknown  quantities,  v,  F,  and  R' .     Solve 


768 


MECHANICS   OF   ENGINEERING. 


(I/)  for  v ;  solve  (2')  for  R'\  substitute  their  values  in  (3') ; 
whence 


h  I/sin 


!  —  cos# 


.(4') 


Since  /"cannot  be  exactly  computed  in  advance,  for  want  of 
knowing  the  value  of  R,  we  calculate  it  approximately  [eq.  (6)> 
§  542b]  for  an  assumed  value  of  R,  insert  it  in  the  above 
equation  (4/),  and  thus  find  an  approximate  value  of  F\  and 
then,  from  (8),  a  corresponding  value  of  R,  from  which  a  new 
value  of  /"can  be  computed.  Thus  after  one  or  two  trials  a 
satisfactory  adjustment  of  dimensions  can  be  secured. 

545.  Variable  Motion.  —  If  a  steady  flow  of  water  of  a  de- 
livery Q,  =  Fv,  =  constant,  takes  place  in  a  straight  open 
channel  the  slope  of  whose  bed  has  not  the  proper  value  to 
maintain  a  "  uniform  motion"  then  "  variable  motion"  ensues 
(the  flow  is  still  steady,  however);  i.e.,  although  the  mean 
velocity  in  any  one  transverse  section  remains  fixed  (with  lapse 
of  time),  this  velocity  has  different  values  for  different  sections  ; 
but  as  the  eq.  of  continuity, 


etc., 


still  holds  (since  the  flow  is  steady),  the  different  sections 

have  different  areas.  If, 
Fig.  616,  a  stream  of 
water  flows  down  an 
inclined  trough  without 
friction,  the  relation 
between  the  velocities 
v0  and  vl  at  any  two 
FIG.  616.  sections  0  and  1  will  be 

the  same  as  for  a  material  point  sliding  down  a  guide  without 

friction  (see  §  79,  latter  part),  viz.  : 


(1) 


VARIABLE   MOTION.       OPEN   CHANNEL.  769 

an  equation  of  heads  (really  a  case  of  Bernoulli's  Theorem, 
§  492).     But,  considering  friction  on  the  bed,  we  must  sub- 

tract the  mean  friction-head  f  -=  ---  —  [see  eqs.  (3)  and  (3'), 

H    2<7 

§  542]  lost  between  0  and  1  ;    this  friction-head  may  also  be 

7  2 

written  thus  :  f  -—•  ~-  ;  and  therefore  eq.  (1)  becomes 


» 


which  is  the  formula  for  variable  motion  ;  and  in  it  I  is  the 
length  of  the  section  considered,  which  should  be  taken  short 
enough  to  consider  the  surface  straight  between  the  end-sec- 
tions, and  the  latter  should  differ  but  slightly  in  area.  The 
subscript  m  may  be  taken  as  referring  to  the  section  midway 
between  the  ends,  so  that  vm2  =  j-(i>02  -f-  v*).  The  wetted  pe- 
ri meter  wm=  i(w.  +  w,),  and  Fm  =  %(FQ  +  F^.  Hence  eq. 
(2)  becomes 


*  _  0  ,         ,.  ,o\ 

BSF  ""%"  ~^+*r  ~^~ 

and  again,  by  putting  v0  —  Q  -=-  FQ  ,  v,  —  Q  -=-  F,  ,  we  may 
write 


F 


whence 


, 
* 


n 


From  eq.  (4),  having  given  the  desired  shapes,  areas,  etc.,  of 
the  end-sections  and  the  volume  of  water,  Q,  to  be  carried  per 
unit  of  time,  we  may  compute  the  necessary  fall,  A,  of  the  sur- 
face, in  length  =  I  ;  while  from  eq.  (5),  having  observed  in  an 
actual  water-course  the  values  of  the  sectional  areas  FQ  and  JF1l  , 
the  wetted  perimeters  w0  and  wl  ,  the  length,  =  I,  of  the  por- 


770 


MECHANICS   OF   ENGINEERING. 


tion  considered,  we  may  calculate  Q  and  thus  gauge  the  stream 
approximately,  without  making  any  velocity  measurements. 

As  to  the  value  of  f,  we  compute  it  from  eq.  (6),  §  542b, 
using  for  R  a  mean  between  the  values  of  the  hydraulic  radii 
of  the  end-sections. 

546.  Bends  in  an  Open  Channel. — According  to  Humphreys 
and  Abbot's  researches  on  the  Mississippi  Kiver  the  loss  of 
head  due  to  a  bend  may  be  put 


, 

^•"536  IT' 


(1) 


in  which  v  must  be  in  ft.  per  sec.,  and  #,  the  angle  ABC,  Fig. 
617,  must  be  in  7r-measure,  i.e.  in  radians. 
The  section  ^must  be  greater  than  100 
sq.  ft.,  and  the  slope  s  less  than  .0008.  v 
is  the  mean  velocity  of  the  water.  Hence 
if  a  bend  occurred  in  a  portion  of  a 
stream  of  length  I,  ©q.  (3)  of  §  542  be- 
comes 


FIG.  617. 


6 


while  eq.  (2)  of  §  545  for  variable  motion  would  then  become 

—  =  -^-  -\-h  —  --^rr-  —  —  -^TT-I  —  -  •  (ft-  and  sec.).  .  (3) 
2#       2#  J1 '  m    Qg       Ooo  7t 

(v  and  d  as  above.)     (For  "  radian"  see  p.  544.) 

547.  Equations  for  Variable  Motion,  introducing  the  Depths. 
— Fig.  618.  The  slope  of  the  bed  being  sin  a  (or  simply  ^, 
Trmeas.),  while  that  of  the  surface  is 
different,  viz., 

sin  fi  =  s  =  h  -4-  /, 
we  may  write 

h  =  d0  -f- 1  sin  a  —  dl , 

FIG.  618. 


VARIABLE   MOTION.       OPEN   CHANNEL.  771 

in  which  d0  and  dl  are  the  depths  at  the  end-sections  of  the 
portion  considered  (steady  flow  with  variable  motion).  With 
these  substitutions  in  eq.  (4),  §  545,  we  have,  solving  for  Z, 


From  which,  knowing  the  slope  of  the  bed  and  the  shape 
and  size  of  the  end-sections,  also  the  discharge  Q,  we  may 
compute  the  length  or  distance,  Z,  between  two  sections  whose 
depths  differ  by  an  assigned  amount  (d0  —  d^.  But  we  can- 
not compute  the  change  of  depth  for  an  assigned  length  Z  from 
(6).  However,  if  the  width  b  of  the  stream  is  constant,  and 
the  same  at  all  depths  ;  i.e.,  if  all  sections  are  rectangles  hav- 
ing a  common  width  ;  eq.  (6)  may  be  much  simplified  by  intro- 
ducing some  approximations,  as  follows  :  We  may  put 


>  '  F? 

._ 
' 


and,  similarly, 
wm 


^    •     1  m  n 

which  approx.  =  ~-    — 
d0b  Zg 

Hence  by  substitution  in  eq.  (6)  we  have 


547a.  Backwater.  —  Let  us  suppose  that   a  steady  flow  las 
been  proceeding  with  uniform  motion  (i.e.,  the  surface  parallel 


772 


MECHANICS    OF    ENGINEERING. 


to  the  bed)  in  an  open  channel  of  indefinite  extent,  and  that  a 
vertical  wall  is  now  set  up  across  the  stream.  The  water  rises 
and  flows  over  the  edge  of  the  wall,  or  weir,  and  after  a  time 
a  steady  flow  is  again  established.  The  depth,  y0  ,  of  the  water 
close  to  the  weir  on  the  up-stream  side  is  greater  than  d0  ,  the 
original  depth.  "We  now  have  "  variable  motion  "  above  the 
weir,  and  at  any  distance  x  up-stream  from  the  weir  the  new 
depth  y  is  greater  than  d0  .  This  increase  of  depth  is  called 
backwater,  and,  though  decreasing  up-stream,  may  be  percep- 
tible several  miles  above  the  weir.  Let  s  be  the  slope  of  the 
original  uniform  motion  (and  also  of  present  bed),  and  v  the 

v* 
velocity   of  the   original   uniform   motion,  and   let    ~k  =  —  . 

y 

Then,  if  the  section  of  the  stream  is  a  shallow  rectangle  of 
constant  width,  we  have  the  following  relation  (Rankine)  : 


0.)],  • 


where  0  is  a  function  of    -,  as  per  following  table  : 


For  —  =  1  .  0 
d0 

<p  =  GO 

1.10 
.680 

1.20 

.480 

1.30 
.376 

1.40 
.304 

1.50 
.255 

1.60 

.218 

1.70 
.189 

For|-  =  1.80 

1.90 

2.00 

2.20 

2.40 

2.60 

2.80 

3.0 

0=  .166 

.147 

.132 

.107 

.089 

.076 

.065 

.056 

00  is  found  from  ^-,  precisely  as  0  from  'J-,  by  use  of  the  table. 

d>0  a0 

"With  this  table  and  eq.  (1),  therefore,  we  can  find  a?,  the  dis- 
tance (u  amplitude  of  backwater")  from  the  weir  of  the  point 
where  any  assigned  depth  y  (or  "  height  of  backwater,"  y  —  d0) 
will  be  found. 

For  example,  Prof.  Bowser  cites  the  case  from  D'Aubuis- 
son's  Hydraulics  of  the  river  Weser  in  Germany,  where  the 
erection  of  a  weir  increased  the  depth  at  the  weir  from  2.5  ft. 
to  10  ft.,  the  flow  having  been  originally  "  uniform"  for  10 
miles.  Three  miles  above  the  dam  the  increase  (y  —  d0)  of 
depth  was  1.25  ft.,  and  even  at  four  miles  it  was  0.75  ft. 


CHAPTER  YIII. 

DYNAMICS  OF  GASEOUS  FLUIDS. 

548.  Steady  Flow  of  a  Gas.—  [KB.  The  student  should  now 
review  §  492  up  to  eq.  (5).]  The  differential  equation  from 
which  Bernoulli's  Theorem  was  derived  for  any  liquid,  with- 
out friction,  was  [eq.  (5),  §  492] 

lrvdv  +  ds  +  -djp  =  0,  .....     (A) 

£/  / 

and  is  equally  applicable  to  the  steady  flow  of  a  gaseous  fluid, 
but  with  this  difference  in  subsequent  work,  that  the  heaviness, 
y  (§  A  °f  the  gas  passing  different  sections  of  the  pipe  or 
stream-line  is,  or  may  be,  different  (though  always  the  same  at 
a  given  point  or  section,  since  the  flow  is  steady).  For  the 
present  we  neglect  friction  and  consider  the  flow  from  a  large 
receiver,  where  the  great  body  of  the  gas  is  practically  at  rest, 
through  an  orifice  in  a  thin  plate,  or  a  short  nozzle  with  a 
rounded  entrance. 

In  the  steady  flow  of  a  gas,  since  y  is  different  at  different 
points,  the  equation  of  continuity  takes  the  form 


Flow  of  weight  per  time-unit  =  Flv1y1  =  F^v^y^  =  etc.  ;  .  (a) 

i.e.,  the  weight  of  gas  passing  any  section,  of  area  F,  per  unit 
of  time,  is  the  same  as  for  any  other  section,  or  Fvy  =  con- 
stant, y  being  the  heaviness  at  the  section,  and  v  the  velocity. 

549.  Flow  through  an  Orifice  —  Remarks.  —  In  Fig.  619  we 
have  a  large  rigid  receiver  containing  gas  at  some  tension,  pn, 
higher  than  that,  pm,  of  the  (still)  outside  air  (or  gas),  and  at 
some  absolute  temperature  Tn,  and  of  some  heaviness  yn\  that  , 
is,  in  a  state  n.  The  small  orifice  of  area  F  being  opened,  the 
gas  begins  to  escape,  and  if  the  receiver  is  very  large,  or  if  the 
supply  is  continually  kept  up  (by  a  blowing-engine,  e.g.),  after 

773 


774 


MECHANICS   OF   ENGINEERING. 


STILL. 
.  AIR.  . 


FIG.  619. 


a  very  short  time  the  flow  becomes  steady.  Let  nm  represent 
any  stream-line  (§  495)  of  the  flow.  According  to  the  ideal 
subdivision  of  this  stream-line  into  JT  ^  ;_  .,, 

laminae  of  equal  mass  or  weight  (not 
equal  volume,  necessarily)  in  estab- 
lishing eq.  (A)  for  any  one  lamina, 
each  lamina  in  the  lapse  of  time  dt 
moves  into  the  position  just  vacated 
by  the  lamina  next  in  front,  and 
assumes  precisely  the  same  velocity, 
pressure,  and  volume  (and  there- 
fore heaviness]  as  that  front  one  had  at  the  beginning  of  the 
dt.  In  its  progress  toward  the  orifice  it  expands  in  volume, 
its  tension  diminishes,  while  its  velocity,  insensible  at  n,  is 
gradually  accelerated  on  account  of  the  pressure  from  behind 
always  being  greater  than  that  in  front,  until  at  m,  in  the 
"  throat"  of  the  jet,  the  velocity  has  become  vm,  the  pressure 
(i.e.,  tension)  has  fallen  to  a  value  pm ,  and  the  heaviness  has 
changed  to  ym.  The  temperature  Tm  (absolute)  is  less  than 
Tn ,  since  the  expansion  has  been  rapid,  and  does  not  depend 
on  the  temperature  of  the  outside  air  or  gas  into  which  efflux 
takes  place,  though,  of  course,  after  the  effluent  gas  is  once 
free  from  the  orifice  it  may  change  its  temperature  in  time. 

We  assume  the  pressure  ^>m  (in  throat  of  jet)  to  be  equal  to 
that  of  the  outside  medium  (as  was  done  with  flow  of  water), 
so  long  as  that  outside  tension  is  greater  than  .527^TC;  but  if  it 
is  less  than  .527  pn  and  is  even  zero  (a  vacuum),  experiment 
seems  to  show  that pm  remains  equal  to  0.527  of  the  interior 
tension  pn\  probably  on  account  of  the  expansion  of  the 
effluent  gas  beyond  the  throat,  Fig.  620,  so 
,///,  that  although  the  tension  in  the  outer  edge, 
^£"  at  a,  of  the  jet  is  equal  to  that  of  the  outside 
medium,  the  tension  at  m  is  greater  because 
of  the  centripetal  and  centrifugal  forces  devel- 
FIO.  620.  oped  in  the  curved  filaments  between  a  and 
m.  (See  §  553.) 

550.  Flow  through  an  Orifice;   Heaviness  assumed  Constant 
during  Flow,  The  Water  Formula. — If  the  inner  tension  pn  ex- 


STEADY   FLOW   OF   GASES.  775 

ceeds  the  outer,  pm  ,  but  slightly,  we  may  assume  that,  like 
water,  the  gas  remains  of  the  same  heaviness  during  flow. 
Then,  for  the  simultaneous  advance  made  by  all  the  laminae  of 
a  stream-line,  Fig.  619,  in  the  time  dt,  we  may  conceive  an 
equation  like  eq.  (A)  written  out  for  each  lamina  between  n 
and  w,  and  corresponding  terms  added  ;  i.e., 

{For  orifices)  .         .    -  fvdv  +  fdz  +  /**&  =  <>.    .  (B) 

qjn  '     t//i  '     Jn        y 

In  general,  y  is  different  in  the.  different  laminae,  but  in  the 
present  case  it  is  assumed  to  be  the  same  in  all  ;  hence,  with 
m  as  datum  level  and  h  —  vertical  distance  from  n  to  w,  we 
have,  from.  eq.  (B), 

^  _  V^_     ,     Q  _  h     ,     Pm  _  Pn  =  Q  „. 

..     fy     *g  r      r 

But  we  may  put  vn  —  0  ;  while  A,  even  if  several  feet,  is 
small  compared  with  —  —  —  .  E.g.,  with  pm  ^  15  Ibs.  per 

sq.  in.  and  pn  =  16  Ibs.  per  sq.  in.,  we  have  for  atmospheric 
air  at  freezing  temperature 


y      y 

Hence,  putting  vn  —  0  and  h  =  0  in  eq.  (1),  we  have 

^m  _  Pn  —Pm  (   Water  formula  ;  for  small  \  /^ 

2g  yn  \  difference  of  pressures,  only.  \ 

The  interior  absolute  temperature  Tn  being  known,  the  yn 
(interior  heaviness)  may  be  obtained  from  yn  =pny*TQ  -r-  Tnp° 
(§  472),  and  the  volume  of  flow  per  unit  of  time  then  obtained 
(first  solving  (2)  for  vm)  is 


(3) 


where  Fm  is  the  sectional  area  of  the  jet  at  m.     If  the  mouth- 
piece or  orifice  has  well-rounded  interior  edges,  as  in  Fig.  541, 


776  MECHANICS    OF   ENGINEERING. 

its  sectional  area  F  may  be  taken  as  the  area  Fm  .  But  if  it  is 
an  orifice  in  "thin  plate,"  putting  the  coefficient  of  contraction 
=  C—  0.60,  we  have 


Fm  =  CF  =  0.60  F-,    and     Qm  =  0.60  Fvm  .     .      (4) 

This  volume,  Qm  ,  is  that  occupied  by  the  flow  per  time-unit 
when  in  state  m,  and  we  have  assumed  that  ym  =  yn  ;  hence 
the  weight  of  flow  per  time-unit  is 

O  =  Qmym  =  FmvmY^  =  Fmvm7n.    .    .    .    (5) 

EXAMPLE.  —  In  the  testing  of  a  blowing-engine  it  is  found 
capable  of  maintaining  a  pressure  of  18  Ibs.  per  sq.  inch  in  a 
large  receiver,  from  whose  side  a  blast  is  steadily  escaping 
through  a  "  thin  plate"  orifice  (circular)  having  an  area  F  =  4 
sq.  inches.  The  interior  temperature  is  20°  Cent,  and  the  out- 
side tension  15  Ibs.  per  sq.  in. 

Required  the  discharge  of  air  per  second,  both  volume  and 
weight.  The  data  are:  pn  =  lS  Ibs.  per  sq.  in.,  Tn  =  %93° 
Abs.  Cent.,  F=  4  sq.  inches,  and  pm  =  15  Ibs.  per  sq.  in.  Use 
ft.-lb.-sec.  system. 

First,  the  heaviness  in  the  receiver  is 


Then,  from  eq.  (2), 

/o  Pn—pm_     /2X32.2[144X18—144X15]_J  555.3 

vm — \  /  ^9 '~  — \ A  AQn —  )      feet 

V  Yn  V  0<089  I  per  sec. 

(97  per  cent  of  this  would  be  more  correct  on  account  of  fric- 
tion.) 

.-.   Qm=F'mvm=.6JFvm  =  TViirX  555.3  =  9.24  cub.  ft.  per  sec. 


at  a  tension  of  15  Ibs.  per  sq.  in.,  and  of  heaviness  (by  hypoth- 
esis) =  .089  Ibs.  per  cub.  ft.     Hence  weight 

=  G  =  9.24  X  .089  =  .82  Ibs.  per  sec. 


FLOW    OF   GASES   BY    MARIOTTE'S   LAW.  777 

The  theoretical  power  of  the  air-compressor  or  blowing-en- 
gine to  maintain  this  steady  flow  can  be  computed  as  in  Exam- 
ple 3,  §  483. 

551.  Flow  through  an  Orifice  on  the  Basis  of  Mariotte's  Law  ; 
or  Isothermal  Efflux.  —  Since  in  reality  the  gas  expands  during 
flow  through  an  orifice,  and  hence  changes  its  heaviness  (Fig. 
619),  we  approximate  more  nearly  to  the  truth  in  assuming 
this  change  of  density  to  follow  Mariotte's  law,  i.e.,  that  the 
heaviness  varies  directly  as  the  pressure,  and  thus  imply  that 
the  temperature  remains  unchanged  during  the  flow.  We 
again  integrate  the  terms  of  eq.  (B\  but  take  care  to  note  that, 
now,  y  is  variable  (i.e.,  different  in  different  laminae  at  the 
same  instant),  and  hence  express  it  in  terms  of  the  variable  p 
(from  eq.  (2),  §  475),  thus  : 


Therefore  the  termy     —  of  eq.  (_Z?)  becomes 


and,  integrating  all  the  terms  of  eq.  (B\  neglecting  A,  and  call- 
ing vn  zero,  we  have 

*V*  _  Pn  i        Pn  (  efflux  ly  Mariotte's  )  /2\ 

%g  ~    Yn          Pm  '      '  \  Law  through  orifice  )  ' 

T       7) 

As  before,  yn  —  ^  f  •  —  y0  ,  and  the  flow  of  volume  per  time- 

*n    Po 

unit  at  m  is 

Qm  =  Fmvm;    .......     (3) 

while  if  the  orifice  is  in  thin  plate,  Fm  may  be  put  =  .60  F^ 
and  the 

weight  of  the  flow  per  time-unit  —  G  =•  FmVmY™.'  •   '(*) 

If  the  mouth-piece  is  rounded,  Fm  —  F=  area  of  exit  orifice 
of  mouth-piece. 


778  MECHANICS   OF   ENGINEERING. 

EXAMPLE. — Applying  eq.  (2)  to  the  data  of  the  example  in 
§  550,  where  yn  was  found  to  be  .089  Ibs.  per  cub.  ft.,  we  have 
[ft.,  lb.,  sec.] 


P 


•%  Qm  =  Fmvm  =  0.60  X  Tf¥  X  584.7  =  9.745  cub.  ft.  per  sec. 
Since  the  heaviness  at  m  is,  from  Mariotte's  law, 

ym  =  SOL  Yn  =  ff  of  .089,     i.e.,     ym  =  .0741  Ibs.  per  cub.  ft., 

Pn 

hence  the  weight  of  the  discharge  is 

G  =  QmYm  =  9.745  X  .0741  =  0.722  Ibs.  per  sec., 

or  about  12  per  cent  less  than  that  given  by  the  "  water  for- 
mula." If  the  difference  between  the  inner  and  outer  tensions 
had  been  less,  the  discrepancy  between  the  results  of  the  two 
methods  would  not  have  been  so  marked. 

552.  Adiabatic  Efflux  from  an  Orifice.  —  It  is  most  logical  to 
assume  that  the  expansion  of  the  gas  approaching  the  orifice, 
being  rapid,  is  adiabatic  (§  478).  Hence  (especially  when  the 
difference  between  the  inner  and  outer  tensions  is  considerable) 
it  is  more  accurate  to  assume  y  as  varying  according  to  Pois- 
son's  Law,  eq.  (1),  §  478  ;  i.e.,  y  =  [yn  +pn*]p*,  in  integrat- 
ing eq.  (B).  Then  the  term 


. 

PJ         ' 


ADIABATIC   FLOW   OF   GASES   THROUGH   ORIFICES.      779 


and  eq.  (^),  neglecting  h  as  before,  and  with  vn  =  0,  becomes 
(See  Fig.  619) 


L  =  l  -      "       -  (Adiabaticflow;  orifice.)  .  (1) 

"9  y^    L-  \Pn'  -J 

Having  observed  ^>n  and  7^  in  the  reservoir,  we  compute 

/y~7 

yn  —  P^°    Q  (from  §  472).     The  gas  at  m,  jnst  leaving  the 

*nj^« 

orifice,  having  expanded  adiabatically  from  the  state  n  to  the 
state  m,  has  cooled  to  a  temperature  Tm  (absolute)  found  thus 

( 


and  is  of  a  heaviness 

« 


and  the  flow  per  second  occupies  a  volume  (immediately  on 
exit) 


and  weighs 

G  =  Fj>mYm  .......    (5) 

EXAMPLE  1.  —  Let  the  interior  conditions  in  the  large  reser- 
voir of  Fig.  619  be  as  follows  (state  n)  :  pn  —  22  -J-  Ibs.  per  sq. 
in.,  and  Tn  =  294°  Abs.  Cent,  (i.e.,  21°  Cent.)  ;  while  ex- 
ternally the  tension  is  15  Ibs.  per  sq.  inch,  which  may  be  taken 
as  being  =  pm  —  tension  at  m,  the  throat  of  jet.  The  opening 
is  a  circular  orifice  in  "  thin  plate"  and  of  one  inch  diameter. 
Required  the  weight  of  the  discharge  per  second  [ft.,  lb.,  sec.; 
g  =  32.2]. 


First,  r.  =  .         X  .0807  =  0.114  Ibs.  per  cub.  ft. 

Then,  from  (1), 

Vm=  /^n-MH  A 

V      r*  L      w  - 

t  _  f  f]  =  8M  ft.  per  sec. 


780  MECHANICS   OF   ENGINEERING. 

Now  F=  \n(-^=  .00546  sq.  ft. 

/.  Qm  =  CFvm  =  60  Fvn  =  0.60  X  .00546  X  844  =  2.765 
cub.  ft.  per  sec.,  at  a  temperature  of 

Tm  =  294  Vf  =  257°  Abs.  Cent.  =  -  16°  Cent, 
and  of  a  heaviness 

ym  =  0.114  V(f7  =  0.085  Ibs.  per  cub.  ft., 
so  that  the  weight  of  flow  per  sec. 

=  O  =  Qmym  =  2.765  X  .085  =  .235  Ibs.  per  sec. 

EXAMPLE  2. — Let  us  treat  the  example  already  solved  by  the 
two  preceding  approximate  methods  (§§  550  and  551)  by  the 
present  more  accurate  equation  of  adiabatic  flow,  eq.  (1). 

The  data  were  (Fig.  619) : 

pn  =  18  Ibs.  per  sq.  in. ;  Tn  =  293°  Abs.  Cent. ; 
pm  =  15       "      "      "     ;  and  F  —  4  sq.  inches 

[/'"being  the  area  of  orifice].     yn  was  found  =  .089  Ibs.  per 
cub.  ft.  in  §  550 ;  hence,  from  eq.  (1), 


18X144 


From  (4), 

Qm  =Fmvm=.6Fvm=.eXr^X^^  =  9.603  cub.  ft.  per  sec.; 

and  since  at  m  it  is  of  a  heaviness 

Ym  =  .089  V(3y  =  .0788  Ibs.  per  cub.  ft, 
we  have  weight  of  flow  per  sec. 

=  G  =  QmYm  =  9.603  X  .0788  =  0.756  Ibs.  per  sec. 


THEORETICAL   MAXIMUM   FLOW   OF    WEIGHT   OF   GAS.      781 

Comparing  the  three  methods  for  this  problem,  we  see  that 

By  the  "  water  formula?  .  .  .   £  —  0.82    Ibs.  per  sec. 
"      isothermal  formula,  .  .  G  =  0.722      "      " 
"      adiabatic formula,    .  .  G  =  0.756 


u        u 


553.  Practical  Notes.  Theoretical  Maximum  Flow  of  Weight. 
—If  in  the  equations  of  §  552  we  write  for  brevity  pm-±-pn= x 
we  derive,  by  substitution  from  (1)  and  (3)  in  (5), 

Weight  of  flow)       n     n  T?   */~5 r-i       m  i 

per  unit  of  time  \  =  G=  Q«r»=F»  Vfypnyn  [l-afl-a*.  .    (1) 

This  function  of  x  is  of  such  a  form  as  to  be  a  maximum  for 
«  =  (j?»-^n)=(|.)s=.512;     ....     (2) 

i.e.,  theoretically,  if  the  state  n  inside  the  reservoir  remains 
the  same,  while  the  outside  tension  (considered  =pm  of  jet, 
Fig.  619)  is  made  to  assume  lower  and  lower  values  (so  that 
a?,  —  pm  -±pn,  diminishes  in  the  same  ratio),  the  maximum  flow 
of  weight  per  unit  of  time  will  occur  when  pm  =  .512  pn,  a 
little  more  than  half  the  inside  tension.  (With  the  more  ac- 
curate value  1.41  (1.408),  instead  of  f,  see  §  478,  we  should 
obtain  .527  instead  of  .512  for  dry  air;  see  §  549.) 

Prof.  Cotterill  says  (p.  544  of  his  "  Applied  Mechanics") : 
"  The  diminution  of  the  theoretical  discharge  on  diminution 
of  the  external  pressure  below  the  limit  just  now  given  is  an 
anomaly  which  had  always  been  considered  as  requiring  ex- 
planation, and  M.  St.  Tenant  had  already  suggested  that  it 
could  not  actually  occur.  In  1866  Mr.  R.  D.  Napier  showed 
by  experiment  that  the  weight  of  steam  of  given  pressure  dis- 
charged from  an  orifice  really  is  independent  of  the  pressure 
of  the  medium  into  which  efflux  takes  place  * ;  and  in  1872 
Mr.  Wilson  confirmed  this  result  by  experiments  on  the  reac- 
tion of  steam  issuing  from  an  orifice." 

"  The  explanation  lies  in  the  fact  that  the  pressure  in  the 

*  When  the  difference  between  internal  and  external  pressures  is  great, — 
should  be  added. 


782  MECHANICS   OF   ENGINEERING. 

centre  of  the  contracted  jet  is  not  the  same  as  that  of  the  sur- 
rounding medium.  The  jet  after  passing  the  contracted  sec- 
tion suddenly  expands,  and  the  change  of  direction  of  the  fluid 
particles  gives  rise  to  centrifugal  forces"  which  cause  the  pres- 
sures to  be  greater  in  the  centre  of  the  contracted  section  than 
at  the  circumference ;  see  Fig.  620. 

Prof.  Cotter-ill  then  advises  the  assumption  ttmt^w=.527pn 
(for  air  and  perfect  gases)  as  the  mean  tension  in  the  jet  at  in 
(Fig.  619),  whenever  the  outside  medium  is  at  a  tension  less 
than  .527j?n .  He  also  says,  u  Contraction  and  friction  must 
be  allowed  for  by  the  use  of  a  coefficient  of  discharge  the 
value  of  which,  however,  is  more  variable  than  that  of  the 
corresponding  coefficient  for  an  incompressible  fluid.  Little  is 
•certainly  known  on  this  point."  See  §§  549  and  554. 

For  air  the  velocity  of  this  maximum  flow  of  weight  is 


r  /7M 

Vd.  of  max.  G  =     997  A  /  ~    ft.  per  sec.,    . 

v    -*•  o— ' 


(3) 


where  Tn  —  abs.  temp,  in  reservoir,  and  TQ  =  that  of  freezing 
point.  Rankine's  Applied  Mechanics  (  p.  584)  mentions  ex- 
periments of  Drs.  Joule  and  Thomson,  in  which  the  circular 
orifices  were  in  a  thin  plate  of  copper  and  of  diameters  0.029 
in.,  0.053  in.,  and  0.084  in.,  while  the  outside  tension  was 
about  one  half  of  that  inside.  The  results  were  84  per  cent 
of  those  demanded  by  theory,  a  discrepancy  due  mainly,  as 
Rankine  says,  to  the  fact  that  the  actual  area  of  the  orifice  was 
used  in  computation  instead  of  the  contracted  section;  i.e.,  con- 
traction was  neglected. 

554.  Coefficients  of  Efflux  by  Experiment.  For  Orifices  and 
Short  Pipes.  Small  Difference  of  Tensions. — Since  the  discharge 
through  an  orifice  or  short  pipe  from  a  reservoir  is  affected 
not  only  by  contraction,  but  by  slight  friction  at  the  edges, 
oven  with  a  rounded  entrance,  the  theoretical  results  for  the 
volume  and  weight  of  flow  per  unit  of  time  in  preceding  para- 
graphs should  be  multiplied  both  by  a  coefficient  of  velocity  0 
and  one  for  contraction  (7,  as  in  the  case  of  water ;  i.e.,  by  a 
coefficient  of  efflux  /*,  =  0(7.  (Of  course,  when  there  is  no 


COEFFICIENTS   OF   EFFLUX.      GAS.  783 

contraction,  C=  1.00,  and  then  ju  =  0  as  with  a  well-rounded 
mouth-piece,  for  instance,  Fig.  541,  and  with  short  pipes.) 

Hence  for  practical  results,  with  orifices  and  short  pipes,  we 
should  write  for  the  weight  of  flow  per  unit  of  time 


Pnl     V  \Pn 

(from  the  equations  of  §  552  for  adiabatic  flow,  as  most  accu- 
rate ;  pm  -T-pn  may  range  from  \  to  1.00).  F=  area  of  orifice,. 
or  of  discharging  end  of  mouth-piece  or  short  pipe.  yn  = 
heaviness  of  air  in  reservoir  and  —  T0pny0  -=-  Tnp0  ,  eq.  (13)  of 
§  437  ;  and  /*  =  the  experimental  coefficient  of  efflux. 

From  his  own  experiments  and  those  of  Koch,  D'Aubuis- 
son,  and  others,  "Weisbach  recommends  the  following  mean 
values  of  ^  for  various  mouthpieces,  when  pn  is  not  more  than 
•J  larger  than  pm  (i.e.,  about  17  %  larger),  for  use  in  eq,  (1)  : 

1.  For  an  orifice  in  a  thin  plate,     .......     /*=0.56 

2.  Forashortcylindricalpipe(innercornersnotrounded),/f==0.75 

3.  For  a  well-rounded  mouth-piece  (like  that  in  Fig.  541),  /*=0.98 

4.  For  a  short  conical  convergent  pipe  (angle  about  6°),  /^=0.92 

EXAMPLE.  —  (Data  from  Weisbach's  Mechanics.)  "If  the 
sum  of  the  areas  of  two  conical  tuyeres  of  a  blowing-machine 
is  F  =.  3  sq.  inches,  the  temperature  in  the  reservoir  15°  Cent., 
the  height  of  the  attached  (open)  mercury  manometer  (see 
Fig.  464)  3  inches,  and  the  height  of  the  barometer  in  the  ex- 
ternal air  29  inches,"  we  have  (ft.,  lb.,  sec.) 

Ab,  Cent.; 


Pn  -  (H)  14.7  X  144  Ibs.  per  sq.  ft.  ; 
yn  =  |||.ff  X  0.0807  =  0.0816  Ibs.  per  cub.  ft, 
while       F=  yf^  sq.  ft.  and  (see  above)  //  =  0.92  ;  hence 


O  =  0.92  X  T|T  (ff)*  V2  X  32.2  X  3X  If  X  14.7  X  144  X  .0816  [1  -  f  f "fl; 


784  MECHANICS   OF  ENGINEEEING. 

i.e..  G  —  .6076  Ibs.  per  second ;  which  will  oc'jupy  a  volume 

y%  =  a  -T-  y0  =  &  -5-.  -0807  =  7.59  cub.  ft. 

at  one  atmosphere  tension  and  freezing-point  temperature; 
while  at  a  temperature  of  Tn  =  288°  Abs.  Cent,  and  tension  of 
pm  —  J-9.  of  one  atmosphere  (i.e.,  in  the  state  in  which  it  was 
on  entering  the  blowing-engine)  it  occupied  a  volume 

v=  iff  -H  X  7.59  =  8.24  cub.  ft. 

(This  last  is  Weisbach's  result,  obtained  by  an  approximate 
formula.) 

555.  Coefficients  of  Efflux  for  Orifices  and  Short  Pipes  for  a 
Large  Difference  of  Tension. — For  values  >  -J  and  <  2,  of  the 
ratio  pn  :  pm,  of  internal  to  external  tension,  Weisbach's  ex- 
periments with  circular  orifices  in  thin  plate,  of  diameters  (=  d) 
from  0.4  inches  to  0.8  inches,  gave  the  following  results : 


Pn  :  Pm 

1.05 

1.09 

1.40 

1.65 

1.90 

2.00 

.  or  d  =  .4in-;  jj.  = 

.55 

.59 

.69 

.72 

.76 

.78 

••    d=.8™-,ju  = 

.56 

.57 

.64 

.68 

.72 

Whence  it  appears  that  ju  increases  somewhat  with  the  ratio  of 
pn  to  pm ,  and  decreases  slightly  for  increasing  size  of  orifice. 

With  short  cylindrical  pipes,  internal  edges  not  rounded, 
and  three  times  as  long  as  wide,  Weisbach  obtained  p,  as 
follows : 


Pn  :  Pm              = 

1.05 

1.10 

1.30 

1.40 

1.70 

1.74 

diam.  =  .4in-  ;  /*  — 

.73 

.77 

.33 

"     =  ,6in-;  n  = 

.81 

.82 

"     =1.0in-  ;  //  = 

.83 

When  the  inner  edges  of  the  0.4  in.  pipe  were  slightly 
rounded,  /*  was  found  =  0.93 ;  while  a  well-rounded  mouth- 
piece of  the  form  shown  in  Fig.  541  gave  a  value  /*  =  from 
.965  to  .968,  for  pn  :  pm  ranging  from  1.25  to  2.00.  These 
values  of  p  are  for  use  in  eq.  (1),  above. 

556.  To  find  the  Discharge  when  the  Internal  Pressure  is 
measured  in  a  Small  Reservoir  or  Pipe,  not  much  larger  than  the 


VELOCITY   OF    APPROACH.      OASES.  785 

Orifice.  —  Fig.  621.     If  the  internal  pressure  pn,  and  tempera- 

ture jTn,  must  be  measured  in  a 

small  reservoir  or  pipe,  n,  whose 

sectional  area  Fn  is  not  very  large    \          = 

compared  with  that  of  the  orifice,   - 

F,  (or  of  the  jet,  Fm  ,)  the  velocity 

vn  at  n  (velocity  of  approach)  can-  FIG.  621. 

not  be  put  =  zero.     Hence,  in  applying  eq.  (E\  §  550,  to  the 

successive  laminae  between  n  and  m,  and  integrating,  we  shall 

have,  for  adiabatic  steady  flow, 


instead  of  eq.  (1)  of  §  552.     But  from  the  equation  of  continuity 
for  steady  flow  of  gases  [eq.  (a)  of  §  548],  Fnvnyn—  Fmvmym  ; 

F  V  2 
hence  v^  —  ^    ™  vm\  while  for  an  adiabatic  change  from  n 

-CnYn 

to  m,  —  =  \~\  ;  whence  by  substitution  in  (1),  solving  for 


vm,  we  have 


\ 

As  before,  from  §§  472  and  478, 


i  __  f  ^V^1 

w  w 


(2) 


*.=  frr-y. (3) 

j^o    ••*•  n 


™*  rm=(*r]r. W 

\Pnl 

Having  observed  pn,  pm,  and  Tn,  tnjn,  and  knowing  the 
area  F  of  the  orifice,  we  may  compute  yn,  ym,  and  vm,  and 
finally  the 

Weight  of  flow  per  time-unit  =  G  =  f^Fvmym^  .    .     (5) 


786  MECHANICS   OF  ENGINEERING. 

taking  /*  from  §  554  or  555.     In  eq.  (2)  it  must  be  remembered 
that  for  an  orifice  in  "  thin  plate,"  Fm  is  the  sectional  area  of 

the  contracted  vein,  and  =  CF\  where  C  may  be  put  =  —  . 

•CM 

EXAMPLE.  —  If  the  diameter  of  AB,  Fig.  621,  is  3£  inches^ 
and  that  of  the  orifice,  well  rounded,  =  2  in.  ;  if  pn  =  1^  at- 
mospheres i=  i  J-  x  14.7  X  144  Ibs.  per  sq.  ft.,  while  pm  =  -^  of 

an  atmos.,  so  that  ^  =  |£,  and    Tn  =  283°  Abs.  Cent.,—  re- 

Pn 

quired  the  discharge  per  second,  using  the  ft.,  lb.,  and  sec. 
From  eq.  (3), 

Yn  =  If  -fit  X  0.0807  =  .08433  Ibs.  per  cub.  ft.  ; 
'whence  (eq.  (4)) 

Ym  —  (f  i-)i/n  =  .07544  Ibs.  per  cub.  ft. 
Then,  from  eq.  (2), 


=  558.1  ft.  per  sec.  ; 


/.    G  =  0.98          558.1  X  .07544  =  .9003  Ibs.  per  sec. 

4V6/ 

557.  Transmission  of  Compressed  Air;  through  very  Long 
Level  Pipes.  Steady  Flow. 

CASE  I.  When  the  difference  between  the  tensions  in  the 
reservoirs  at  the  ends  of  the  pipe  is  small.  —  Fig.  622.  Under 


FIG.  622. 


these  circumstances  it  is  simpler  to  employ  the  form  of  formula 
that  would  be  obtained  for  a  liquid  by  applying  Bernoulli's 
Theorem,  taking  into  account  the  "  loss  of  head  "  occasioned 


TRANSMISSION   OF  COMPKESSED  AIR.  787 

by  the  friction  on  the  sides  of  the  pipe.  Since  the  pipe  is 
very  long,  and  the  change  of  pressure  small,  the  mean  velocity 
in  the  pipe,  v',  assumed  to  be  nearly  the  same  at  all  points 
along  the  pipe,  will  not  be  large  ;  hence  the  difference  be- 
tween the  velocity-heads  at  n  and  m  will  be  neglected  ;  a  cer- 
tain mean  heaviness  yf  will  be  assigned  to  all  the  gas  in  the 
pipe,  as  if  a  liquid. 

Applying  Bernoulli's  Theorem,  with  friction,  §  516,  to  the 
ends  of  the  pipe,  n  and  m,  we  have  (as  for  a  liquid) 

^.^,   0  =  ^  +  ^  +  0-  4/1^-.          (1) 

W^?  fy     rf  J<*fy 

Putting  (as  above  mentioned)  vm*  —  vn*  —  0,  we  have,  more 
simply, 


The  value  of  f  as  coefficient  of  friction  for  air  in  long 
pipes  is  found  to  be  somewhat  smaller  than  for  water  ;  see  next 
paragraph. 

558.  Transmission  of  Compressed  Air.  Experiments  in  the  St. 
Gothard  Tunnel,  1878.—  [See  p.  96  of  Yol.  24  (Feb.  '81),  Yan 
Nostrand's  Engineering  Magazine.]  In  these  experiments, 
the  temperature  and  pressure  of  the  flowing  gas  (air)  were  ob- 
served at  each  end  of  a  long  portion  of  the  pipe  which  delivered 
the  compressed  air  to  the  boring-machines  three  miles  distant 
from  the  tunnel's  mouth.  The  portion  considered  was  selected 
at  a  distance  from  the  entrance  of  the  tunnel,  to  eliminate  the 
fluctuating  influence  of  the  weather  on  the  temperature  of  the 
flowing  air.  A  steady  flow  being  secured  by  proper  regulation 
of  the  compressors  and  distributing  tubes,  observations  were 
made  of  the  internal  pressure  (p\  internal  temperature  (T),  as 
well  as  the  external,  at  each  end  of  the  portion  of  pipe  con- 
sidered, and  also  at  intermediate  points  ;  also  of  the  weight 
of  flow  per  second  G  =  Q0y0,  measured  at  the  compressors 
under  standard  conditions  (0°  Cent,  and  one  atmos.  tension). 
Then  knowing  the  p  and  T  at  any  section  of  the  pipe,  tha 


788 


MECHANICS    OF   ENGINEERING. 


heaviness  y  of  the  air  passing  that  section  can  be  computed 

r          y        T>    T  ~\ 
from  —  -  =  — - .  — °-     and  the  velocity  v  =  G  -=-  Fy,  F  being 

L          Yo      Po    -L  — ' 

the  sectional  area  at  that  point.  Hence  the  mean  velocity  v', 
and  the  mean  heaviness  7',  can  be  computed  for  this  portion 
of  the  pipe  whose  diameter  =  d  and  length  —  I.  In  the  ex- 
periments cited  it  was  found  that  at  points  not  too  near  the 
tunnel-mouth  the  temperature  inside  the  pipe  was  always 
about  3°  Cent,  lower  than  that  of  the  tunnel.  The  values  of 
/in  the  different  experiments  were  then  computed  from  eq. 
(2)  of  the  last  paragraph  ;  i.e., 


>n—  Pm  _ 
Y' 


7      «./a 


all  the  other  quantities  having  been  either  directly  observed, 
or  computed  from  observed  quantities. 

THE  ST.  GOTHARD  EXPERIMENTS. 

[Concrete  quantities  reduced  to  English  units.] 


No. 

l 

(feet.) 

d 

(ft.) 

(Ibs.  cub. 

ft.) 

Atmospheres. 

Pn  ~  Pm 
Ibs.  sq.  in. 

v' 
ft.  per  sec. 

mean 
temp. 
Cent. 

Pn 

Pm 

1 

15092 

| 

0.4058 

5.60 

5.24 

5.29 

19.32 

21° 

.0035 

2 

15092 

} 

0.3209 

4.35 

4.13 

3.23 

16.30 

21° 

.0038 

8 

15092 

.2803 

3.84 

3.65 

2.79 

15.55 

21° 

.0041 

4 

1712 

* 

.3765 

5.24 

5.00 

3.52 

37.13 

26.5 

.0045 

0 

1712 

.  3009 

4.13 

4.06 

1.03 

30.82 

26.5 

.0024(?) 

6 

1712 

* 

.2641 

3.65 

3.54 

1.54 

29.34 

26.5 

.0045 

In  the  article  referred  to  (Yan  Nostrand's  Mag.)  f  is  not 
computed.  The  writer  contents  himself  with  showing  that 
"Weisbach's  values  (based  on  experiments  with  small  pipes  and 
high  velocities)  are  much  too  great  for  the  pipes  in  use  in  the 
tunnel. 

With  small  tubes  an  inch  or  less  in  diameter  "Weisbach 
found,  for  a  velocity  of  about  80  ft.  per  second,/* =.0060; 
for  still  higher  velocities/  was  smaller,  approximately,  in  ac- 
cordance with  the  relation 

f—  .0542 

yV(in  ft.  per  sec.) 


TRANSMISSION   OF   COMPRESSED   AIR.  789 

On  p.  370,  vol.  xxiv,  Yan  Nostrand's  Mag.,  Prof.  Kobinson 
of  Ohio  mentions  other  experiments  with  large  long  pipes. 

From  the  St.  Gothard  experiments  a  value  off=  .004  may 
be  inferred  for  approximate  results  with  pipes  from  3  to  8  in. 
in  diameter. 

EXAMPLE.  —  It  is  required  to  transmit,  in  steady  flow,  a  supply 
of  G  —  6.456  Ibs.  of  atmospheric  air  per  second  through  a  pipe 
30000  ft.  in  length  (nearly  six  miles)  from  a  reservoir  where 
the  tension  is  6.0  atmos.  to  another  where  it  is  5.8  atmos.,  the 
mean  temperature  in  the  pipe  being  80°  Fahr.,  =  24°  Cent. 
(i.e.  —  297°  Abs.  Cent.).  Kequired  the  proper  diameter  of 
pipe  ;  d  =  ?  The  value  .  /=  .00425  will  be  used,  and  the  ft- 
Ib.-sec.  system  of  units.  The  mean  volume  passing  per  second 
in  the  pipe  is 

<X=Q*Y'  ........     (3) 

Q>  Q> 

The  mean  velocity  may  thus  be  written  :  vf  =  -~  =  -—  ;  .   (4) 

The  mean  heaviness  of  the  flowing  air,  computed  for  a  mean 
tension  of  5.9  atmospheres,  is,  by  §  472, 


r'=  -~  •         X  .0807  =  0.431  Ibs.  per  cub.  ft.  ; 
and  hence,  see  eq.  (3), 

• 


at  tension  of  5.9  atmos.,  and  temperature  297°  Abs.  Cent. 
Now,  from  eq.  (2), 


whence 

J/  y'l 

~ 


790  MECHANICS   OF   ENGINEERING. 

and  hence,  numerically, 

5  /  4  X  .00425  X  0.431  X  30000  X  (14.74)a 
~  Y  (.7854)*[14.7  X  144(6.00  -  5.80)]2  X  32.2  ~ 

559.  (Case  II  of  g  557)  Long  Pipe,  with  Considerable  Differ- 
ence of  Pressure  at  Extremities  of  the  Pipe.  Flow  Steady. — Fig. 
623.  If  the  difference  between  the  end-tensions  is  compara- 
tively great,  we  can  no  longer  deal  with  the  whole  of  the  air 


2£ 

•  —  p~                      --^ 

*-fl 

/*£> 

:::A|      iB—V:i  :-.•/.> 

n  .  '  .  •  •   ""•'•  "? 

—*<-->• 

ds 
FIG.  623. 

in  the  pipe  at  once,  as  regards  ascribing  to  it  a  mean  velocity 
and  mean  tension,  but  must  consider  the  separate  laminae, 
such  as  AE  (a  short  length  of  the  air-stream)  to  which  we  may 
apply  eq.  (2)  of  §  556 ;  A  and  E  corresponding  to  the  n  and 
m  of  Fig.  622.  Since  t\\Q  pn—  pm,  I,  yf,  and  v'  of  §  559 
correspond  to  the  —  dp,  ds,  y,  and  v  of  the  present  case  (short 
section  or  lamina),  we  may  write 

r  (I'D  <?)* 

_<*P_  =  4:fVcls (1) 

y          '  dty 

But  if  G  =  weight  of  flow  per  unit  of  time,  we  have  at  any 
section,  Fvy  =  G  (equation  of  continuity)  ;  i.e.,  v  —  G  -r-  Fy, 
whence  by  substitution  in  eq.  (1)  we  have 

dp  _ 


Eq.  (2)  contains  three  variables,  y,  p,  and  s  (=  distance  of 
lamina  from  n').  As  to  the  dependence  of  the  heaviness  y  on 
the  tension^?  in  different  laminae,  experiment  shows  that  in  most 
cases  a  uniform  temperature  is  found  to  exist  all  along  the 
pipe,  if  properly  buried,  or  shaded  from  the  sun  ;  the  loss  of 
heat  by  adiabatic  expansion  being  in  great  part  made  up  by 
the  heat  generated  by  the  friction  against  the  walls  of  the 


GAS   IN    LONG   PIPES.       LARGE   FALL   OF   TENSION. 


pipe.  This  is  due  to  the  small  loss  of  tension  per  unit  of 
length  of  pipe  as  compared  with  that  occurring  in  a  short  dis- 
charge pipe  or  nozzle.  Hence  we  may  treat  the  flow  as  iso- 
t/iermal,  and  write  p  -^-  y  —2}n'^-  Yn'  (§  4^5,  Mariotte's  Law). 

Hence  y  —  —  ^>,  which   substituted  in  eq.  (2)  enables  us  to 

Pn' 

write:  T 

-pdp  =         ; 


Performing  the  integration,  noting  that  at  nr p  =pn>,  s  —  0, 
and  at  mf  p  =  pm>  and  s  =  I,  we  have 

ir.»  »  _     >  ,n  -  ^fl    &   Pn>         j  isothermal  flow  \  ,.. 

*UV       pm  \    -tyd'jr*"^,'       (    in  long  pipes    ] 

It  is  here  assumed  that  the  tension  at  the  entrance  of  the  pipe 
is  practically  equal  to  that  in  the  head  reservoir,  and  that  at 
the  end  (ra')  to  that  of  the  receiving  reservoir;  which  is  not 
strictly  true,  especially  when  the  corners  are  not  rounded.  It 
will  be  remembered  also  that  in  establishing  eq.  (2)  of  §  556 
(the  basis  of  the  present  paragraph),  the  "inertia"  of  the  gas 
was  neglected ;  i.e.,  the  change  of  velocity  in  passing  along 
the  pipe.  Hence  eq.  (4)  should  not  be  applied  to  cases  where 
the  pipe  is  so  short,  or  the  difference  of  end-tensions  so  great, 
as  to  create  a  considerable  difference  between  the  velocities  at 
the  two  ends  of  the  pipe. 

EXAMPLE. — A  well  or  reservoir  supplies  natural  gas  at  a  ten- 
sion of  pn>  =  30  Ibs.  per  sq.  inch.  Its  heaviness  at  0°  Cent, 
and  one  atmosphere  tension  is  .0484  Ibs.  per  cub.  foot.  In 
piping  this  gas  along  a  level  to  a  town  two  miles  distant,  a 
single  four-inch  pipe  is  to  be  employed,  and  the  tension  in  the 
receiving  reservoir  (by  proper  regulation  of  the  gas  distributed 
from  it)  is  to  be  kept  equal  to  16  Ibs.  per  sq.  in.  (which  would 
sustain  a  column  of  water  about  2  ft.  in  height  in  an  open 
water  manometer,  Fig.  465). 


792  MECHANICS    OF  ENGINEERING. 

The  mean  temperature  in  the  pipe  being  17°  Cent.,  required 
the  amount  (weight)  of  gas  delivered  per  second,  supposing 
leakage  to  be  prevented  (formerly  a  difficult  matter  in  practice). 
Solve  (4)  for  G,  and  we  have 


First,  from  §  472,  with  Tn,  =  Tm,  =  290°  Abs.  Cent.,  we 
compute 


273 
Hence  with/  —  .005, 


x  -30  x 


4X.005  X  10560  x  46454: 
=  0.337  Ibs.  per  sec. 

(For  compressed  atmospheric  air,  under  like  conditions,  we 
would  have  G  =  0.430  Ibs.  per  second.) 

Of  course  the  proper  choice  of  the  coefficient/"  has  an  im- 
portant influence  on  the  result. 

From  the  above  result  (G  =  0.337  Ibs.  per  second)  we  can 
compute  the  volume  occupied  by  this  quantity  of  gas  in  the 

fi- 

receiving  reservoir,  using  the  relation  Qm»  =  —  . 

Ym! 

The  heaviness  ym>  of  the  gas  in  the  receiving  reservoir  is 
most  easily  found  from  the  relation  -^—  =  ±-^  .which  holds 

Ym'  Yn> 

good    since    the    flow    is    isothermal.     I.e.,  &2*'=  46454  ft.; 

Ym' 

whence  ym>  =  0.049  Ibs.  per  cubic  foot,  pm>  being  16  X  144 
Ibs.  per  sq.  ft. 
Hence 


FLOW   OF   GAS   IK   PIPES.  793 

It  should  be  said  that  the  pressure  at  the  up-stream  end  of 
the  pipe  depends  upon  the  rate  of  flow  allowed  to  take  place. 

With  no  flow  permitted,  the  pressure  in  the  tube  of  a  gas- 
well  has  in  some  cases  reached  the  high  figure  of  500  or  600 
Ibs.  per  sq.  in.  . 

560.  Rate  of  Decrease  of  Pressure  along  a  Long  Pipe, — Con- 
sidering further  the  case  of  the  last  paragraph,  that  of  a 
straight,  long,  level  pipe  of  uniform  diameter,  delivering  gas 
from  a  storage  reservoir  into  a  receiving  reservoir,  we  note 
that  if  in  eq.  (4)  we  retain  pm>  to  indicate  the  tension  in  the 
receiving  reservoir,  but  let  pn>  denote  in  turn  the  tension  at 
points  in  the  pipe  successively  further  and  further  (a  distance 
x]  from  the  receiving  reservoir  w',  we  may  write  x  for  I  and 
obtain  the  equation  (between  two  variables,  pn>  and  x) 

Pn?  —Pn/  =  Const.  X  ®. (6) 

This  can  be  used  to  bring  out  an  interesting  relation  men- 
tioned by  a  writer  in  the  Engineering  News  of  July  1887 
(p.  71),  viz.,  the  fact  that  in  the  parts  of  the  pipe  more  distant 
from  the  receiving  end,  m',  the  distance  along  the  pipe  in 
which  a  given  loss  of  pressure  occurs  is  much  greater  than 
near  the  receiving  end. 

To  make  a  numerical  illustration,  let  us  suppose  that  the 
pipe  is  of  such  size,  in  connection  with  other  circumstances, 
that  the  tension  pn>  at  A,  a  distance  x  =  six  miles  from  m',  is 
two  atmospheres,  the  tension  in  the  receiving  reservoir  being 
one  atmosphere ;  that  is,  that  the  loss  of  tension  between  A 
and  m!  is  one  atmosphere.  If  we  express  tensions  in  atmos- 
pheres and  distances  in  miles,  we  have  for  the  value  of  the 
constant  in  eq.  (6),  for  this  case, 

Const.  =  (4  —  1)  -4-  6  —  f ;   (for  assumed  units.}  .   .  (7) 

Now  let pn>  =  the  tension  at  B,  a  point  18  miles  from  m', 
and  we  have,  from  eqs.  (6)  and  (7),  the  tension  at  B  —  3.1  f> 
atmospheres.  Proceeding  in  this  manner,  the  following  set  of 
values  is  obtained  : 


794 


MECHANICS   OF   ENGINEERING. 


Point. 

Total  distance 
from  m'. 

Distance  be- 
tween consecu- 
tive points. 

Tension  at 
point. 

Loss  of  ten- 
sion in  each  in- 
terval. 

F 
E 
D 

C 
B 
A 
m' 

126  mi 
90 
60 
36 
18 
6 
0 

les. 

36  mi 
30 
24 

"18 
12 
6 

les. 

8.00  at 
6.78 
5.56 
4.35 
3.16 
2.00 
1.00 

m. 

1.22  at 
1.22 
1.21 
1.19 
1.16 
1.00 

m. 

If  the  distances  and  tensions  in  the  second  and  fourth 
columns  be  plotted  as  abscissae  and  ordinates  of  a  curve,  the 
latter  is  a  parabola  with  its  axis  following  the  axis  of  the  pipe ; 
its  vertex  is  not  at  m',  however. 

561.  Long  Pipe  of  Variable  Diameter, — Another  way  of  stat- 
ing the  fact  mentioned  in  the  last  paragraph  is  as  follows  :  At 
the  up-strearn  end  of  the  pipe  of  uniform  diameter  the  gas  is 
of  much  greater  density  than  at  the  other  extremity  (the 
heaviness  is  directly  as  the  tension,  the  temperature  being  as- 
sumed the  same  throughout  the  pipe),  and  the  velocity  of  its 
motion  is  smaller  than  at  the  discharging  end  (in  the  same 
ratio).  It  is  true  that  the  frictional  resistance  per  unit  of 
length  of  pipe  varies  directly  as  the  heaviness  [eq.  (1),  §  510], 
but  also  true  that  it  varies  as  the  square  of  the  velocity ;  so 
that,  for  instance,  if  the  pressure  at  a  point  A  is  double  that 
at  B  in  the  pipe  of  constant  diameter,  it  implies  that  the 
heaviness  and  velocity  at  A  are  double  and  half,  respectively, 
those  at  B,  and  thus  the  gas  at  A  is  subjected  to  only  half  the 
frictional  resisting  force  per  foot  of  length  as  compared  with 
that  at  B.  Hence  the  relatively  small  diminution,  per  unit  of 
length,  in  the  tension  at  the  up-stream  end  in  the  example  of 
the  last  paragraph. 

In  the  pipe  of  uniform  diameter,  as  we  have  seen,  the  greater 
part  of  the  length  is  subjected  to  a  comparatively  high  ten- 
sion, and  is  thus  under  a  greater  liability  to  loss  by  leakage 
than  if  the  decrease  of  tension  were  more  uniform.  The 
total  "hoop-tension"  (§  426)  in  a  unit  length  of  pipe,  also,  is 
proportional  to  the  gas  tension,  and  thinner  walls  might  be 
employed  for  the  down-stream  portions  of  the  pipe  if  the  gas 


FLOW   OF   GAS   IN   PIPES.  795 

tension  in  those  portions  could  be  made  smaller  than  as  shown 
in  the  preceding  example. 

To  secure  a  more  rapid  fall  of  pressure  at  the  up-stream  end 
of  the  pipe,  and  at  the  same  time  provide  for  the  same  delivery 
of  gas  as  with  a  pipe  of  uniform  diameter  throughout,  a  pipe 
of  variable  diameter  may  be  employed,  that  diameter  being 
considerably  smaller  at  the  inlet  than  that  of  the  uniform  pipe 
but  progressively  enlarging  down-stream.  This  will  require 
the  diameters  of  portions  near  the  discharging  end  to  be  larger 
than  in  the  uniform  pipe,  and  if  the  same  thickness  of  metal 
were  necessary  throughout,  there  would  be  no  saving  of  metal, 
but  rather  the  reverse,  as  will  be  seen ;  but  the  diminished 
thickness  made  practicable  in  those  parts  from  a  less  total  hoop 
tension  than  in  the  corresponding  parts  of  the  uniform  pipe 
more  than  compensates  for  the  extra  metal  due  to  increased 
circumference,  aside  from  the  diminished  liability  to  leakage, 
which  is  of  equal  importance. 

A  simple  numerical  example  will  illustrate  the  foregoing. 
The  pipe  being  circular,  we  may  replace  F  by  ^nd*  in  equation 
{4),  and  finally  derive,  G  being  given, 

i—  7  — li 

-     (8) 


Let  A  be  the  head  reservoir,  and  mf  the  receiving  reservoir, 
and  B  a  point  half-way  between.  At  A  the  tension  is  10  at- 
mospheres ;  at  m',  2  atmospheres.  For  transmitting  a  given 
weight  of  gas  per  unit-time,  through  a  pipe  of  constant  diam- 
ter  throughout,  that  diameter  must  be  (tensions  in  atmospheres  ; 
2/0  being  the  length),  by  eq.  (8), 

d  =  67!  *=       -0208*  =  °-46      -  •    8 


If  we  substitute  for  the  pipe  mentioned,  another  having  a  con- 
stant diameter  dl  from  A  to  B,  where  we  wish  the  tension  to 
be  5  atmospheres,  and  a  different  constant  diameter  d%  from  B 
to  m',  we  derive  similarly 


100  1- 


796  MECHANICS   OF   ENGINEERING. 

and 


It  is  now  to  be  noted  that  the  sum  of  d1  and  d^  is  slightly 
greater  than  the  double  of  d  ;  so  that  if  the  same  thickness  of 
metal  were  used  in  both  designs  the  'compound  pipe  would 
require  a  little  more  material  than  the  uniform  pipe;  but, 
from  the  reasoning  given  at  the  beginning  of  this  paragraph, 
that  thickness  may  be  made  considerably  less  in  the  down- 
stream part  of  the  compound  pipe,  and  thus  economy  secured. 

[In  case  of  a  cessation  of  the  flow,  the  gas  tension  in  the 
whole  pipe  might  rise  to  an  equality  with  that  of  the  head- 
reservoir  were  it  not  for  the  insertion,  at  intervals,  of  auto- 
matic regulators,  each  of  which  prevents  the  decrease  of  ten- 
sion on  its  down-stream  side  below  a  fixed  value.  To  provide 
for  changes  of  length  due  to  rise  and  fall  of  temperature,  the 
pipe  is  laid  with  slight  undulations.] 

It  is  a  noteworthy  theoretical  deduction  that  a  given  pipe  of 
variable  diameter  connecting  two  reservoirs  of  gas  at  specified 
pressures  will  deliver  the  same  weight  of  gas  as  before,  if 
turned  end  for  end.  This  follows  from  equation  (3)',  §  559. 
With  d  variable,  (3)'  becomes  (with  F—  \ittf) 


S* 

, 
***' 


, 

-.   .   (9) 


„„ 

''**-€?' 

(C"  is  a  constant.) 

f*m'  ds 
But  Jn,    -'-  is  evidently  the  same  in  value  if  the  pipe  be 

turned  end  for  end.  In  commenting  on  this  circumstance,  we 
should  remember  (see  §  559)  that  the  loss  of  pressure  along  the 
pipe  is  ascribed  entirely  to  frictional  resistance,  and  in  no  de- 
gree to  changes  of  velocity  (inertia). 

On  p.  73  of  the  Engineering  News  of  July  1887  are  given 
the  following  dimensions  of  a  compound  pipe  in  actual  use, 
and  delivering  natural  gas.  The  pressure  in  the  head-reservoir 
is  319  Ibs.  per  sq.  in.;  that  in  the  receiving  reservoir,  65.  For 
2.84  miles  from  the  head-reservoir  the  diameter  of  the  pipe  is 


NATURAL   GAS.       COEFFICIENT   OF   FLUID    FRICTION.     797 


8  in. ;  throughout  the  next  2.75  miles,  10  in.*,  while  in  the 
remaining  3.84  miles  the  diameter  is  12  in.  At  the  two 
points  of  junction  the  pressures  are  stated  to  be  185  and  132 
Ibs.  per  sq.  in.,  respectively,  during  the  flow  of  gas  under  the 
conditions  mentioned. 

561a.  Values  of  the  Coefficient  of  Fluid  Friction  for  Natural 
Gas. — In  the  Ohio  Keport  on  Economic  Geology  for  1888  may 
be  found  an  article  by  Prof.  S.  W.  Robinson  of  the  University 
of  that  State  describing  a  series  of  interesting  experiments 
made  by  him  on  the  flow  of  natural  gas  from  orifices  and 
through  pipes.  By  the  insertion  of  Pitot  tubes  approximate 
measurements  were  made  of  the  velocity  of  the  stream  of  gas 

«/  C5 

in  a  pipe.  The  following  are  some  of  the  results  of  these  ex- 
periments, p1  —p^  representing  the  loss  of  pressure  (in  Ibs.  per 
sq.  inch)  per  mile  of  pipe-length,  and /the  coefficient  of  fluid 
friction,  in  experiments  with  a  six-inch  pipe  : 


Pi  -Pi 

1.00 

1.50 

2.25 

2.50 

5.75 

6.25 

f 

0.0025 

0.0037 

0.0052 

0.0059 

0.0070 

0.0060 

In  the  flow  under  observation  Prof.  Robinson  concluded 
that  f  could  be  taken  as  approximately  proportional  to  the 
fourth  root  of  the  cube  of  the  velocity  of  flow  ;  though  calling 
attention  to  the  fact  that  very  reliable  results  could  hardly  be 
expected  under  the  circumstances. 


CHAPTEK  IX. 


IMPULSE  AND  RESISTANCE  OF  FLUIDS. 

562.  The  so-called  "Reaction"  of  a  Jet  of  Water  flowing  from 
a  Vessel. — In  Fig.  624,  if  a  frictionless  but  water-tight  plug  B 

be  inserted  in  an  orifice  in  the 
vertical  side  of  a  vessel  mounted 
on  wheels,  the  resultant  action  of 
the  water  on  the  rigid  vessel  (as  a 
whole)  consists  of  its  weight  6r, 
and  a  force  P'  —  Fhy  (in  which 
F=  the  area  of  orifice)  which  is 
the  excess  of  the  horizontal  hydro- 
static pressures  on  the  vessel  wall 
toward  the  right  ( ||  to  paper)  over 
those  toward  the  left,  since  the 

pressure  P,  =  Fhy,  exerted  on  the  plug  is  felt  by  the  post  (7, 
and  not  by  the  vessel.     Hence  the  post  D  receives  a  pressure 


P'  =  Fhy. 


(1) 


Let  the  plug  B  be  removed.  A  steady  flow  is  then  set  up 
through  the  orifice,  and  now  the  pressure  against  the  post  D  is 
2Fhy  (as  will  be  proved  in  the  next  paragraph) ;  for  not  only 
is  the  pressure  Fhy  lacking  on  the  left,  because  of  the  orifice, 
but  the  sura  of  all  the  horizontal  components  (  ||  to  paper)  of. 
the  pressures  of  the  liquid  filaments  against  the  vessel  wall 
around  the  orifice  is  less  than  its  value  before  the  flow  began, 
by  an  amount  Fhy.  A  resistance  R  =  %Fhy  being  provided, 
and  the  post  removed,  a  slow  uniform  motion  may  be  main- 
tained toward  the  right,  the  working  force  being  %Fhy  =  P" 

798 


KE  ACTION"    OF   A   JET. 


FIG.  625. 


(see  Fig.  625 ;  R  is  not  shown).     If  an  insufficient  resistance 
be  furnished  before  removing  the  post  D,  .... ....  ^ 

the  vessel  will  begin  to  move  toward  the 
right  with  an  acceleration,  which  will 
disturb  the  surface  of  the  water  and 
change  the  value  of  the  horizontal  force. 
This  force 

is  called  the  "  reaction'*  of  the  water-jet ; 
y  is  the  heaviness  of  the  liquid  (§  7). 

Of  course,  as  the  flow  goes  on,  the 
water  level  sinks  and  the  '•  reaction"  diminishes  accordingly. 
Looked  upon  as  a  motor,  the  vessel  may  be  considered  to  be  a 
piston-less  and  valve-less  water-pressure  engine,  carrying  its 
own  reservoir  with  it. 

In  Case  II  of  §  500  we  have  already  had  a  treatment  of  the 
"  Reaction-wheel "  or  "  Barker's  mill,"  which  is  a  practical 
machine  operating  on  this  principle,  an.d  will  be  again  con- 
sidered in  "Notes  on  Hydraulic  Motors." 

563.  "  Reaction"  of  a  Liquid  Jet  on  the  Vessel  from  which  it 
Issues. — Instead  of  showing  that  the  pressures  on  the  vessel 
close  to  the  orifice  are  less  than  they  were  when  there  was  no 
flow  by  an  amount  Fhy  (a  rather  lengthy  demonstration), 
another  method  will  be  given,  of  greater  simplicity  but  some- 
what fanciful. 

If  a  man  standing  on  the  rear  platform  of  a  car  is  to  take  up 
in  succession,  from  a  basket  on  the  car,  a  number  of  balls  of 
equal  mass  —  M,  and  project  each  one  in  turn  horizontally 
backward  with  an  acceleration  —  p,  he  can  accomplish  this 
only  by  exerting  against  each  ball  a  pressure  =  Mp,  and  in  the 
opposite  direction  against  the  car  an  equal  pressure  =  -Mp.  If 
this  action  is  kept  up  continuously  the  car  is  subjected  to  a 
constant  and  continuous  forward  force  of  P"  —  Mp. 

Similarly,  the  backward  projection  of  the  jet  of  water  in  the 
case  of  the  vessel  at  rest  must  occasion  a  forward  force  against 
the  vessel  of  a  value  dependent  on  the  fact  that  in  each  small 
interval  of  time  At  a  small  mass  AM  of  liquid  has  its  velocity 
changed  from  zero  to  a  backward  velocity  of  v  =  V%gh ;  that 


800  MECHANICS    OF   ENGINEERING. 

is,  has  been  projected  with  a  mean  acceleration  of  p  = 
so  that  the  forward  force  against  the  vessel  is 


P"  =  mass  X  ace.  =  --  .....     (3) 

At 

If  Q  =  the  volume  of  water  discharged  per  unit  time,  then 

Ov 
AM.  =  -££  At,  and  since  also  Q  —  Fv  —  FV%gh,  eq.    (3)  be- 

•/  &/i^ 

comes  "Reaction"  of  jet  =  P"  —  %Fhy.     .    .     .     (4) 

(A  similar  proof,  resulting  in  the  same  value  for  P"  ,  is 
easily  made  if  the  vessel  has  a  uniform  motion  with  water  sur- 
face horizontal.) 

If  the  orifice  is  in  "  thin  plate,"  we  understand  by  F  the 
area  of  the  contracted  section.  Practically,  we  have  v  =  0  V%g"h> 
(§  495),  and  hence  (3)  reduces  to 

P"  =  %<pFhy.        .     .)'  ..I....     (5) 

Weisbach  mentions  the  experiments  of  Mr.  Peter  Ewart  of 
Manchester,  England,  as  giving  the  result  P"  =  \.1SFliy 
with  a  well-rounded  orifice  as  in  Fig.  625.  He  also  found 
<t>  =  .94  for  the  same  orifice,  so  that  by  eq.  (4)  we  should  have 

P"  =  ^(MJFhy  =  \MFhy. 

With  an  orifice  in  thin  plate  Mr.  Ewart  found  P"  — 
"L.^Fhy.  As  for  a  result  from  eq.  (4),  we  must  put,  for  F\ 
the  area  of  the  contracted  section  .647^  (§  495),  which,  with 
0  =  .96,  gives 

P"  =  2(.96)2  .  §±Fhy  =  LlSFhy.     ...  .    (6) 

Evidently  both  results  agree  well  with  experiment. 

Experiments  made  by  Prof.  J.  B.  Webb  at  the  Stevens 
Institute  (see  Journal  of  the  Franklin  Inst.,  Jan.  '88,  p.  35) 
also  confirm  the  foregoing  results.  In  these  experiments  the 
vessel  was  suspended  on  springs  and  the  jet  directly  down- 
ward,  so  that  the  "reaction"  consisted  of  a  diminution  of  the 
tension  of  the  springs  during  the  flow. 

564.  Impulse  of  a  Jet  of  Water  on  a  Fixed  Curved  Vane  (with 
Borders).  —  The  jet  passes  tangentially  upon  the  vane.  Fig. 


IMPULSE   OF  JET. 


801 


626.  B  is  the  stationary  nozzle  from  which  a  jet  of  water  of 
cross-section  F  (area)  and  velocity  =  c  impinges  tangentially 
upon  the  vane,  which  has 
plane  borders,  parallel  to 
paper,  to  prevent  the  lat- 
eral escape  of  the  jet. 
The  curve  of  the  vane  is 
not  circular  necessarily. 
The  vane  being  smooth, 
the  velocity  of  the  water 
in  its  curved  path  remains 

x  < 

FIG.  626. 


—  c  at  ail  points  a^ong 
the  curve.  Conceive  the 
curve  divided  into  a  great 
number  of  small  lengths 


dP 


each  =  ds,  and  subtending  some 
angle  =  d<p  from  its  own  centre  of  curvature,  its  radius  of 
curvature  being  =  r  (different  for  different  <fo's),  which  makes 
some  angle  =  0  with  the  axis  I^(~|  to  original  straight  jet 
J3A).  At  any  instant  of  time  there  is  an  arc  of  water  AD  in 
contact  with  the  vane,  exerting  pressure  upon  it.  The  pres- 
sure dP  of  any  ds  of  the  vane  against  the  small  mass  of  water 
Fds  .  y  -r-  g  then  in  contact  with  it  is  the  "  deviating"  or  "  cen- 
tripetal "  force  accountable  for  its  motion  in  a  curve  of  radius 
—  r,  and  hence  must  have  a  value 


(§76) 


(1) 


The  opposite  and  equal  of  this  force  is  the  dP  shown  in 
Fig.  614,  and  is  the  impulse  or  pressure  of  this  small  mass 
against  the  vane.  Its  ^-component  is  dX  =  dP  sin  0.  By 
making  0  vary  from  0  to  <*,  and  adding  up  the  corresponding 
values  of  dXt  we  obtain  the  sum  of  the  ^"-components  of  the 
small  pressures  exerted  simultaneously  against  the  vane  by  the 
arc  of  water  then  in  contact  with  it  ;  i.e.,  noting  that  ds=rd& 


/»*  =  «        /»« 

/.  /  dX—  I  dP  .  sin  0  = 
v  ^° 


802 


MECHANICS   OF   ENGINEERING. 


hence  the  X-impulse  )       Fye  n  -.      Qyc  ri  .,   /0, 

•     ,  /?      T  -^         }•  = — —  1 — cos  a  =  — —  1 — cos  a  L  (2) 
against  fixed  vane    )          ^,  ^ 

in  which  Q  =  Fc  =  volume  of  water  which  passes  through  the 
nozzle  (and  also  =  that  passing  over  the  vane,  in  this  case)  per 
unit  of  time,  and  a  —  angle  between  the  direction  of  the 
stream  leaving  the  vane  (i.e.,  at  D)  and  its  original  direction 
(BA  of  the  jet)  ;  i.e.,  a  —  total  angle  of  deviation.  Similarly, 
the  sum  of  the  ^-components  of  the  dP's  of  Fig.  626  may  be 
shown  to  be 


/*«  Q<vc 

Y -impulse  on  fixed  vane  =  JQ  dP  .  cos  0  =  ¥1—  sin  or...(2)/ 


Hence  the  resultant  impulse  on  the  vane  is  a  force 


P"  = 


Y  2  = 


;,    .  .  (3> 


and  makes  such  an  angle  <*',  Fig.  627,  with  the  direction 
that 

,        Y  sin  a 

tan  &'  =  ---=  —- 


. 
X        1  —  cos  a 


FIG.  62« 


FIG.  628. 


For  example,  if  OL  =  90°,  then  a?  =  45°;  while  if  a  =  180°, 
Fig.  628,  we  have  a'  =  0° ;  i.e.,  P"  is  parallel  to  the  jet 
and  its  value  is 


IMPULSE   OF   JET   ON   VANE. 


803 


565.  Impulse  of  a  Jet  on  a  Fixed  Solid  of  Revolution  whose 
Axis  is  Parallel  to  the  Jet. — If  the 
curved  vane,  with  borders,  of  the  pre- 
ceding paragraph  be  replaced  by  a 
solid  of  revolution,  Fig.  629,  with  its 
axis  in  line  of  the  jet,  the  resultant 
pressure  of  the  jet  upon  it  will  simply 
be  the  sum  of  the  X-components  (i.e., 
=  to  BA)  of  the  pressures  on  all  ele- 
ments of  the  surface  at  a  given  instant ;  i.e., 


FIG.  629. 


(5) 


while  the  components  1  to  X,  all  directed  toward  the  axis  of 
the  solid,  neutralize  each  other.  For  a  fixed  plate,  then,  Fig. 
630,  at  right  angles  to  the  jet,  we  have  for  the  force,  or  "  im- 
pulse" (with  a  =  90°), 


The   experiments  of  Bidone,  made   in 
1838,  confirm   the  truth  of  eq.  (6)  quite 
closely,  as  do  also  those  of  two  students  of 
FIG. 630.  -          the  University  of  Pennsylvania  at  Phila- 
delphia (see  Jour,  of  the  Frank.  Inst.  for  Oct.  '87,  p.  258). 

Eq.(6)  is  applicable  to  the  theo- 
ry of  Pitot's  Tube  (see  §  539), 
Fig.  631,  if  we  consider  the  edge 
of  the  tube  plane  and  quite  wide.  •'.'.••:•'•.'• :-'':':.'-.'-.:-:': V.B'- 
The  water  in  the  tube  is  at  rest, 
and  its  section  at  A  (of  area— F) 
may  be  treated  as  a  flat  vertical 
plate  receiving  not  only  the 
hydrostatic  pressure  Fxy,  due 
to  the  depth  x  below  the  sur-  -  FIG.  63i. 

face,  but  a  continuous  impulse  P"  —  Fc*y  -f-  g  [see  eq.  (6)]. 


804 


MECHANICS   OF   ENGINEERING. 


For  the  equilibrium  of  the  end  A,  of  the  stationary  column 
AD,  we  most  have,  therefore, 


Fxy  + 


'n    i.e,    A'  =  (2.0).  .(7) 


The  relation  in  equation  (7)  corresponds  reasonably  well 
with  the  results  of  Weisbaeh's  experiments  with  the  instru- 
ment mentioned  in  §  539.  Pitot  himself,  on  trial  of  an  in- 
strument in  which  the  edges  of  the  tube  at  A  were  made  flar- 
ing or  conically  divergent,  like  a  funnel,  found 


(7)' 


while  Darcy,  desirous  that  the  end  of  the  tube  should  occasion 
as  little  disturbance  as  possible  in  the  surrounding  stream, 
made  the  extremity  small  and  conically 
convergent.  The  latter  obtained  the 
relation 

h'  =  almost  exactly  (1.0)  —  .  .  (7)" 

i/ 

(See  §  539.) 

If  the  solid  of  revolution  is  made  cup- 
shaped,  as  in  Fig.  632,  we  have  (as  in 
Fig.  628)  a  —  180°,  and  therefore,  from 

eq.  (5),   ' 


FIG.  632. 


EXAMPLE.  —  Fig.  632.  If  c  =  30  ft.  per  sec.  and  the  jet 
(cylindrical)  has  a  diameter  of  1  inch,  the  liquid  being  water, 
so  that  y  =  62.5  Ibs.  per  cub.  ft.,  we  have  [ft.,  lb.,  sec.] 


the  impulse  (force)  =  P"  = 


32.2 


=  19.05  Ibs. 


Experiment  would  probably  show  a  smaller  result. 


IMPULSE  OF  JET  ON   MOVING  VANE. 


803 


o  - 


FIG.  633. 


566.  Impulse  of  a  Liquid  Jet  upon  a  Moving  Vane  having 
Lateral  Borders  and  Moving  in  the  Direction  of  the  Jet. — Fig. 
633.  The  vane  has  a  motion  of  translation  (§  108)  in  the 
same  direction  as  the  jet.  Call  this  the  axis  X.  It  is  moving 
with  a  velocity  v  away  from  the  jet  (or,  if  toward  the  jet,  v 
is  negative).  "We  con- 
sider v  constant,  its  ac- 
celeration being  prevented 
by  a  proper  resistance 
(such  as  a  weight  =  G) 
to  balance  the  ^-com- 
ponents of  the  arc-pres- 
sures. Before  coming  in 
contact  with  the  vane, 
which  it  does  tangentially 
(to  avoid  sudden  devia- 
tion), the  absolute  velocity 
(§  83)  of  the  water  in  the 
jet  =  <?,  while  its  velocity 
relatively  to  the  vane  at  A  is  =  c  —  v\  and  it  will  now  be 
proved  that  the  relative  velocity  along  the  vane  is  constant. 
See  Fig.  634.  Let  v  —  the  velocity  of  the  vane  (of  each 
point  of  it,  since  its  motion  is  one  of  translation),  and  u  =  the 
velocity  of  a  water  particle  (or  small  mass  of  water  of  length 
=  ds)  relatively  to  the  point  of  the  vane  which  it  is  passing. 
Then  w,  the  absolute  velocity  of  the  small  mass,  is  the  diago- 
nal formed  on  u  and  v.  Neglecting  friction,  the  only  actual 
force  acting  on  the  mass  is  P,  the  pressure  of  the  vane  against 
it,  and  this  is  normal  to  the  curve.  Now  an  imaginary  system 
of  forces,  equivalent  to  this  actual  system  of  one  force  P,  i.e., 
capable  of  producing  the  same  motion  in  the  mass,  may  be 
conceived  of,  consisting  of  the  individual  forces  which  would 
produce,  separately,  the  separate  motions  of  which  the  actual 
motion  of  this  small  mass  M  is  compounded.  These  com- 
ponent motions  are  as  follows : 

1.  A  horizontal  uniform  motion  of  constant  velocity  =u; 
and 

2.  A  motion  in  the  arc  of  a  circle  of  radius  =  r  and  with  a 


806  MECHANICS    OF   ENGINEEEING. 

velocity  =  u,  which  we  shall  consider  variable  until  proved 
otherwise. 

Motion  1  is  of  such  a  nature  as  to  call  for  no  force  (by  New- 
ton's tirst  law  of  motion),  while  motion  2  could  be  maintained 

li/T     2 

by  a  system  of  two  forces,  one  normal,  Pn ,  =  -  — ,  and  the 

other  tangential,  Pt  —  M  —  [see  eq.  (5),  p.  76].     This  imagi- 

d-t 

nary  system  of  forces  is  shown  at  (II.),  Fig.  634,  and  is  equiv- 


(i.) 


FIG.  C34. 


alent  to  the  actual  system  at  (I.).  Therefore  2  (tang,  com- 
pons.)  in  (I.)  should  be  equal  to  2  (tang,  compons.)  in  (II.) ; 
whence  we  have 

Pt  =  0;    i.e.,    M^  =  0',     or    ~  =  0;    .     .     (1) 
ctt  dt 

i.e.,  u  is  constant  along  the  vane  and  is  equal  to  c  —  v  at  every 
point.  (The  weight  of  the  mass  has  been  neglected  since  the 
height  of  the  vane  is  small.)  In  Fig.  634  the  symbol  w0  has 
been  used  instead  of  c,  and  the  point  0  corresponds  to  A  in 
Fig.  633. 

[N.B.  If  the  motion  of  the  vane  were  rotary,  about  an  axis 
"I  to  AB  (or  to  c\  this  relative  velocity  would  be  different  at 
different  points.  See  Notes  on  Hydraulic  Motors.  If  the 
radius  of  motion  of  the  point  A,  however,  is  quite  large  com- 
pared with  the  projection  of  AD  upon  this  radius,  the  relative 
velocity  is  approximately  =  c  —  v  at  all  parts  of  the  vane, 
and  will  be  taken  =  c  —  v  in  treating  the  "  Hurdy-gurdy"  in 
§  567.] 


WORK   OF   JET   ON  VANE.  807 

By  putting  ^(normal  corapons.)  of  (I.)  =  2  (normal  com- 
pons.)  in  (II.)  we  have 


p  _.  P  .  p  _  M      -         - 

*  —  ±*n,    i.e.,    r  -JXL--        —  --  ,    .     .     .     (Z) 

so  that  to  find  the  sura  of  the  ^-components  of  the  pressures 
exerted  against  the  vane  simultaneously  by  all  the  small  masses 
of  water  in  contact  with  it  at  any  instant,  the  analysis  differs 
from  that  in  §  564  only  in  replacing  the  c  of  that  article  by 
the  (c  —  v)  of  this.  Therefore 

2(X-pressures)  =  PX  =  —  ^  (c  —  ^)2[1  —  cos  a],    .    (3) 

t/ 

(where  a  is  the  angle  of  total  deviation,  relatively  to  vane,  of 
the  stream  leaving  the  vane,  from  its  original  direction),  and 
is  seen  to  be  proportional  to  the  square  of  the  relative  velocity. 
F\§  the  sectional  area  of  jet,  and  y  the  heaviness  (§  T)  of  the 
liquid.  The  Z"-component  (or  PY)  of  the  resultant  impulse 
is  counteracted  by  the  support  EF,  Fig.  633.  Hence,  for  a 
uniform  motion  to  be  maintained,  with  a  given  velocity  =  v, 
the  weight  G  must  be  made  =  Px  of  eq.  (3).  (We  here 
neglect  friction  and  suppose  the  jet  to  preserve  a  practically 
horizontal  direction  for  an  indefinite  distance  before  meeting 
the  vane.  If  this  uniform  motion  is  to  be  toward  the  jet,  v 
will  be  negative  in  eq.  (3),  making  Px  (and  .*.  G)  larger  than 
for  a  positive  v  of  same  numerical  value. 

As  to  the  doing  of  work  [§§  128,  etc.],  or  exchange  of 
energy,  between  the  two  bodies,  jet  and  vane,  during  a  uni- 
form motion  away  from  the  jet,  Px  exerts  a  power  of 


-vMl-cosa],  .     .     .     (4) 

in  which  Z  denotes  the  number  of  units  of  work  done  per  unit 
of  time  by  Px\  i.e.,  the  power  (§  130)  exerted  by  Px. 
If  v  is  negative,  call  it  —  v\  and  we  have  the 

Power  expended  )        D    /       Fy  ,     .     /VI  /ri  n        /KN 

%  vane  upon  jet  \  =  P°°v  :    ~f  ('  +  ")"  t1  ~  CO8  «]•  •    (5) 


808 


MECHANICS   OF  ENGINEERING. 


Of  course,  practically,  we  are  more  concerned  with  eq.  (4) 
than  with  (5).  The  power  L  in  (4)  is  a  maximum  for  v  =  -Jc; 
but  in  practice,  since  a  single  moving  vane  or  float  cannot 
utilize  the  water  of  the  jet  as  fast  as  it  flows  from  the  nozzle, 
let  us  conceive  of  a  succession  of  vanes  coming  into  position 
consecutively  in  front  of  tlie  jet,  all  having  the  same  velocity 
v ;  then  the  portion  of  jet  intercepted  between  two  vanes  is  at 
liberty  to  finish  its  work  on  the  front  vane,  while  additional 
work  is  being  done  on  the  hinder  one ;  i.e.,  the  water  will  be, 
utilized  as  fast  as  it  issues  from  the  nozzle. 

"With  such  a  series  of  vanes,  then,  we  may  put  Q',  =  Fc, 
the  volume  of  flow  per  unit  of  time  from  the  nozzle,  in  place 
of  f(c  _  v)  —  the  volume  of  flow  per  unit  of  time  over  the 
vane,  in  eq.  (4)  ;  whence 


Power  exerted  on  \  _ 
series  of  vanes     j 


—  jt-L  [i  _  cos  a~\(c  —  v)v. 

i/ 


Making  v  variable,  and  putting  dLf-^-dv=Oy  whence  c— 20=0,. 
we  find  that  for  v  =  \c,  L ' ,  the  power,  is  a  maximum.  As- 
suming different  values  for  a,  we  find  that  for  a  =  180°,  i.e., 
by  the  use  of  a  semicircular  vane,  or  of  a  hemispherical  cup, 
Fig.  635,  with  a  point  in  middle,  1  —  cos  a  is  a  max.,  =  2 ; 

whence,  with  v  =  %c,  we  have,  as   the- 

maximum  power, 


•  Tf       — 

7     4^  max.  — ' 


in  which  M '  denotes  the  mass  of  the  flow 
per   unit   of  time   from   the    stationary 

nozzle.     Now is  the  entire  'kinetic 

energy  furnished  per  unit  of  time  by  the 
jet ;  hence  the  motor  of  Fig.  635  (series 
FIG.  635.  of  cups)  has  a  theoretical  efficiency  of 

unity,  utilizing  all  the  kinetic  energy  of  the  water.  If  this  is 
true,  the  absolute  velocity  of  the  particles  of  liquid  where  they 
leave  the  cup,  or  vane,  should  be  zero,  which  is  seen  to  be  true,, 


POWER   OF   JETS. 


809 


as  follows :     At  H,  or  H ',  the  velocity  of  the  particles  rela- 
tively to  the  vane  is  =  c  —  v  =  what  it  was  at  A,  and  hence 

s*  s* 

is  =  'c  —  —  =  — ;    hence  at  H  the  absolute  velocity  is  w  = 

/%  /* 

(rel.  veloc.  —  toward  left)— (veloc.  —  of  vane  toward  right)  =  0 ; 

2  2 

Q.E.D.     For  v  >  or  <  \c  this  efficiency  will  not  be  attained. 

567.  The  California  " Hurdy-gurdy  ;"  or  Pelton  Wheel,—  The 
efficiency  of  unity  in  the  series  of  cups  just  mentioned  is  in 
practice  reduced  to  80  or  85  per  cent  from  friction  and  lateral 
escape  of  water.  The 
Pelton  wheel  or  Cali- 
fornia "  Hurdy-gur- 
dy," shown  (in  prin- 
ciple only)  in  Fig.  636, 
is  designed  to  utilize 
the  mechanical  rela- 


tion    just    presented, 

and  yields  results  con-  £ TQis^Ms 

firming  the  above  the-  ' ^ 

ory,  viz.,  that  with  the 

linear  velocity  of  the  FIG.  636. 

s* 

cup-centres  regulated  to  equal  — ,  and  with  a  =  180°,  the  effi- 

2 

ciency  approaches  unity  or  100  per  cent.  Each  cup  has  a  pro- 
jecting sharp  edge  or  rib  along  the  middle,  to  split  the  jet ;  see 
Fig.  635. 

This  wheel  was  invented  to  utilize  small  jets  of  very  great 
velocities  (c)  in  regions  just  deserted  by  "  hydraulic  mining" 
operators.  Although  c  is  great,  still,  by  giving  a  large  value 

s* 

to  r,  the  radius  of  the  wheel,  the  making  of  v  =  -  does  not 

2 

necessitate  an  inconveniently  great  speed  of  rotation  (i.e., 
revolutions  per  unit  of  time).  The  plane  of  the  wheel  may 
be  in  any  convenient  position. 

In  the  London  Engineer  of  May  '84,  p.  397,  is  given  an  ac- 
count of  a  test  made  of  a  "  Hurdy-gurdy,"  in  which  the  motor 


810 


MECHANICS    OF   ENGINEERING. 


showed  an  efficiency  of  87  per  cent.  The  diameter  of  the 
wheel  was  only  6  ft.,  that  of  the  jet  1.89  in.,  and  the  head  of 
the  supply  reservoir  386  ft.,  the  water  being  transmitted 
through  a  pipe  of  22  inches  diameter  and  6900  ft.  in  length. 
107  H.  P.  was  developed  by  the  wheel. 

EXAMPLE.- — If  the  jet  in  Fig.  636  has  a  velocity  c  =  60  ft. 
per  second,  and  is  delivered  through  a  2-inch  nozzle,  the  total 
power  due  to  the  kinetic  energy  of  the  water  is  (ft.,  lb.,  sec.) 


t/ 

and  if,  by  making  the  velocity  of  the  cups  =  -  =  30  ft.  per 

sec.,  85  per  cent  of  this  power  can  be  utilized,  the  DO^ver  of 
the  wheel  at  this  most  advantageous  velocity  is 

L  =  .85  X  4566.9  =  3881  ft.  Ibs.  per  sec.  =  7.05  horse-power 

[since  3881  -r-  550  =  7.05]  (§  132).  For  a  cup-velocity  of  30 
ft.  per  sec.,  if  we  make  the  radius,  ?•,  =  10  feet,  the  angular 
velocity  of  the  wheel  will  be  GO  =  v  -z-  r  =  3.0  radians  per 
sec.  (for  radian  see  Example  in  §  428 ;  for  angular  velocity, 
§  110),  which  nearly  =  ?r,  thus  implying  nearly  a  half-revolu- 
tion per  sec. 

568.  Oblique  Impact  of  a  Jet  on  a  Moving   Plate  having 

no  Border. — The  plate 
has  a  motion  of  trans- 
lation with  a  uniform 
D  veloc.  —  v  in  a  direc- 
tion parallel  to  jet, 
whose  velocity  is  =  c. 
At  0  the  filaments  of 
liquid  are  deviated,  so 
that  in  leaving  the  plate 
their  particles  are  all 
found  in  the  moving 
plane  BB'  of  the  plate 
surface,  but  the  respective  absolute  velocities  of  these  particles 


FIG.  637. 


OBLIQUE   JET  AND   PLATE.  811 

depend  on  the  location  of  the  point  of  the  plate  where  they 
leave  it,  being  found  by  forming  a  diagonal  on  the  relative 
veloc.  c  —  v  and  the  velocity  v  of  the  plate.  For  example,  at 
B  the  absolute  velocity  of  a  liquid  particle  is 


w  =  BE  =  Vv*  +  (o  —  v)*  +  2v(c  —  v)  cos  a, 
while  at  B'  it  is 


w'  =  B'E'  —  1/V  +  (c  —  v)'*  —  Zv(c  —  v)  cos  a ; 

but  evidently  the  component  1  to  plate  (the  other  component 
being  parallel)  of  the  absolute  velocities  of  all  particles  leaving 
the  plate,  is  the  same  and  =  v  sin  a.  The  skin-friction  of  the 
liquid  on  the  plate  being  neglected,  the  resultant  impulse  of 
the  jet  against  the  plate  must  be  normal  to  its  surface,  and  its 
amount,  jP,  is  most  readily  found  as  follows : 

Denoting  by  AM  the  mass  of  the  liquid  passing  over  the 
plate  in  a  short  time  At,  resolve  the  absolute  velocities  of  all 
the  liquid  particles,  before  and  after  deviation,  into  com- 
ponents 1  to  the  plate  (call  this  direction  Y)  and  ||  to  the 
plate.  Before  meeting  the  plate  the  particles  composing  AM 
have  a  velocity  in  the  direction  of  Y  of  cv  =  c  sin  a ;  on  leav- 
ing the  plate  a  velocity  in  direction  of  Y  of  v  sin  a  :  they  have 
therefore  lost  an  amount  of  velocity  in  direction  of  Y  = 
(c  —  v)  sin  a  in  time  At ;  i.e.,  they  have  suffered  an  average 
retardation  (or  negative  acceleration)  in  a  Indirection  of 

(  neg.  accelera-  )     _  (c  —  v)  sin  a  ,_ 

*V?:  jtiou  ||  to  Y  f  :         ~~ZT 

Hence  the  resistance  in  direction  of  Y(i.e.,  the  equal  and  op- 
posite of  P  in  figure)  must  be 

PY  =  mass  X  I^-accel.  = (c—  v)  sin  a ;  .     .     (2) 

At 

and  therefore,  since =  M  =  -—  =  mass  of  liquid  passing 

At  g 


812  MECHANICS   OF   ENGINEEKING. 

over  the  plate  per  unit  of  time  (not  that  issuing  from  nozzle), 
we  have 


Impulse  orpres-  )  =p=  Qy(c_v)^a  =  ^r(c_vysin     (3) 
sure  on  plate     j  g  ^  g  ^ 

in  which  F  =  sectional  area  of  jet  before  meeting  plate. 

[N.B.  Since  eq.  (3)  can  also  be  written  P  =.  Me  sin  a  — 
Mv  sin  or,  and  Me  sin  a  may  be  called  the  I^-momentum  before 
contact,  while  Mv  sin  a  is  the  JT-momentum  after  contact  (of 
the  mass  passing  over  plate  per  unit  of  time),  this  method  may 
be  said  to  be  founded  on  the  principle  of  momentum  which  is 
nothing  more  than  the  relation  that  the  accelerating  force  in 
any  direction  =  mass  X  acceleration  in  that  direction  ;  e.g.,, 
Px  =  Mpx;  Py  =  Mpy',  see§74.] 

If  we  resolve  P,  Fig.  637,  into  two  components,  one,  P',  \\ 
to  the  direction  of  motion  (v  and  <?),  and  the  other,  P'1  ',  ~|  to 
the  same,  we  have 


and 

P"  =  P  cos  a  =  52-  (e  —  v)  sin  a  cos  a.     .     .     (5) 

($  =  T^c  —  v)  —  volume  passing  over  the  plate  per  unit  of 
time.)  The  force  P"  does  no  work,  while  the  former,  P' ^ 
does  an  amount  of  work  P'v  per  unit  of  time ;  i.e.,  exerts  a 
power  (one  plate) 

=  L  —  P'v  =  Q^(c  —  v)v  sin3  a.  (6) 

ff 

If,  instead  of  a  single  plate,  a  series  of  plates,  forming  a 
regular  succession,  is  employed,  then,  as  in  a  previous  paragraph, 
we  may  replace  Q,  =  F(c  —  v\  by  Qf  =  Fc,  obtaining  as  the 

Power  exerted  ly  jet )  =  z/  =  ^V  (   _   }     .  _ 

<?7i  ^/•^^  of  plates     \  g     ^ 


IMPULSE   OF   JET   ON   OBLIQUE  PLATE. 


For  v  =  G-  and  a  =  90°  we  have 

2 


=i^yfL  =i^v  ,ft 

o  o"o       o         *       *      *      *      *       \   / 


=  only  half  the  kinetic  energy  (per  time-unit)  of  the  jet. 

569.  Rigid  Plates  Moving  in  a  Fluid,  Totally  Submerged. 
Fluid  Moving  against  a  Fixed  Plate.  Impulse  and  Resistance. — 
If  a  thin  flat  rigid  plate  have  a  motion  of  uniform  translation 
with  velocity  =  v  through  a  fluid 
which  completely  surrounds  it,  Fig.  "- — - 
638,  a  resistance  is  encountered  (which  ^^ 
must  be  overcome  by  an  equal  and  op-  - 
posite  force,  not  shown  in  figure,  to 
preserve  the  uniform  motion)  consist- 
ing of  a  normal  component  JV,  1  to 
plate,  and  a  (small)  tangential  com- 
ponent, or  skin-friction,  T^  \\  to  plate.  FIG.  ess. 
Unless  the  angle  a,  between  the  surface  of  plate  and  the  direc- 
tion of  motion  O  ...  -y,  is  very  small,  i.e.  unless  the  plate  i& 
moving  nearly  edgewise  through  the  fluid,  N  is  usually  much 
greater  than  T.  The  skin-resistance  between  a  solid  and  a  fluid 
has  already  been  spoken  of  in  §  510. 

When  the  plate  and  fluid  are  at  rest  the  pressures  on  both 
sides  are  normal  and  balance  each  other,  being  ordinary  static 
fluid  pressures.  When  motion  is  in  progress,  however,  the 
normal  pressures  on  the  front  surface  are  increased  by  the 
components,  normal  to  plate,  of  the  centrifugal  forces  of  the 
curved  filaments  (such  as  AB)  or  "stream-lines,"  while  on 
the  back  surface,  D,  the  fluid  does  not  close  in  fast  enough  to 
produce  a  pressure  equal  to*  that  (even)  of  rest.  In  fact,  if  the 
motion  is  sufficiently  rapid,  and  the  fluid  is  inelastic  (a  liquid), 
a  vacuum  may  be  maintained  behind  the  plate,  in  which  case 
there  is  evidently  no  pressure  on  that  side  of  the  plate. 

Whatever  pressure  exists  on  the  back  acts,  of  course,  to- 
diminish  the  resultant  resistance.  The  water  on  turning  the 
sharp  corners  of  the  plate  is  broken  up  into  eddies  forming  a, 


£14  MECHANICS   OF   ENGINEERING. 

"  wake"  behind.  From  the  accompaniment  of  these  eddies, 
the  resistance  in  this  case  (at  least  the  component  N  normal  to 
plate)  is  said  to  be  due  to  "eddy-making /"  though  logically 
Ave  should  say,  rather,  that  the  body  does  not  derive  the  assist- 
ance (or  negative  resistance)  from  behind  which  it  would  ob- 
tain if  eddies  were  not  formed  ;  i.e.,  if  the  fluid  could  close  in 
behind  in  smooth  curved  stream-lines  symmetrical  with  those 
in  front. 

The  heat  corresponding  to  the  change  of  temperature  pro- 
duced in  the  portion  of  fluid  acted  on,  by  the  skin-friction 
and  by  the  mutual  friction  of  the  particles  in  the  eddies,  is  the 
equivalent  of  the  work  done  (or  energy  spent)  by  the  motive 
force  in  maintaining  the  uniform  motion  (§  149).  (Joule's 
experiments  to  determine  the  Mechanical  Equivalent  of  Heat 
were  made  with  paddles  moving  in  water.) 

If  the  fluid  is  sea-water,  the  results  of  Col.  Beaufoy's  ex- 
periments are  applicable,  viz.: 

The  resistance,  per  square  foot  of  area,  sustained  by  a  sub- 
merged plate  moving  normally  to  itself  [i.e.,  a  =  90°]  in  sea- 
water  with  a  velocity  of  v  =  10  ft.  per  second  is  112  Ibs.  He 
also  asserts  fchatjftw  other  velocities  the  resistance  varies  as  the 
square  of  the  velocity.  This  latter  fact  we  would  be  led  to 
suspect  from  the  results  obtained  in  §  568  for  the  impulse  of 
jets;  also  in  §565  [see  eq.  (6)].  Also,  that  when  the  plate 
moved  obliquely  to  its  normal  (as  in  Fig.  638)  the  resistance 
was  nearly  equal  to  (the  resistance,  at  same  velocity,  when 
a  =  90°)  X  (the  sine  of  the  angle  a}  ;  also,  that  the  depth  of 
submersion  had  no  influence  on  the  resistance. 

Confining  our  attention  to  a  plate  moving  nor- 
mally to  itself,  Fig.  639,  let  F  =  area  of  plate, 
y  =  heaviness  (§  409)  of  the  fluid,  v  —  the  uni- 
form velocity  of  plate,  and  g  =  the  acceleration 
of   gravity   (=  32.2   for   the   foot   and    second ; 
=  9.81  for  the  metre  and  second).     Then  from 
the  analogy  of  eq.  (6),  §  565,  where  velocity  c  of 
the  jet  against  a  stationary  plate  corresponds  to 
FIG.  639.        the  velocity  v  of   the  plate  in  the  present  case 
moving  through  a  fluid  at  rest,  we  may  write 


-ITY 


PLATES  MOVING  IN  FLUIDS. 

Resistance  of  fluid  \  _  T>_  ^p    ^          \  v  normal )          ,^ 
to  moving  plate     }  ~         ~  ^    ? '  %g  '     '  (   to  plate  j  *  *  "  W 

And  similarly  for  the  impulse  of  an  indefinite  stream  of  fluid 
against  a  fixed  plate  (  ~|  to  velocity  of  stream),  v  being  the 
velocity  of  the  current, 

Impulse  of  current }  __  p r'//7    ^          j  ^  normal )         /o\ 

upon  fixed  plate    )  ~  ^     ^  2^~  '  '  '  (   to  plate  f  *  V*   * 

The  2g  is  introduced  simply  for  convenience ;  since,  having 
v  given,  we  may  easily  find  v*  -=-  %g  from  a  table  of  velocity- 
heads  ;  and  also  (a  ground  of  greater  importance)  since  the  co- 
efficients £  and  C'  which  depend  on  experiment  are  evidently 
abstract  numbers  in  the  present  form  of  these  equations  (for 
It  and  P  are  forces,  and  Fyv*  -r-  %g  is  the  weight  (force)  of 
an  ideal  prism  of  fluid ;  hence  C  and  C,'  must  be  abstract 
numbers.) 

From  Col.  Beaufoy's  experiments  (see  above),  we  have  for 
sea-water  [ft.,  lb.,  sec.],  putting  R  =  112  Ibs.,  F  =  1  sq.  ft.,, 
y  =  64  Ibs.  per  cub.  ft.,  and  v  =  10  ft.  per  second, 

2  X  32.2  X  112  =       3 
"  1.0  X  64  X  102 

Hence  in  eq.  (1)  for  sea-water,  we  may  put  C  =  1.13  (with 
y  =  64  Ibs.  per  cub.  ft.). 

From  the  experiments  of  Dubuat  and  Thibault,  "Weisbach 
computes  that  for  the  plate  of  Fig.  639,  moving  through  either 
water  or  air,  C  =  1-25  for  eq.  (1),  in  which  the  y  for  air  must 
be  computed  from  §  437 ;  while  for  the  impulse  of  water  or 
air  on  fixed  plates  he  obtains  C'  —  1.86  for  use  in  eq.  (2).  It 
is  hardly  reasonable  to  suppose  that  C  and  C'  should  not  be 
identical  in  value,  and  Prof.  Unwin  thinks  that  the  difference 
in  the  numbers  just  given  must  be  due  to  errors  of  experi- 
ment. The  latter  value,  C'  =  1-86,  agrees  well  with  equation 
(6)  below.  For  great  velocities  C  and  C'  are  greater  for  air 
than  for  water,  since  air,  being  compressible,  is  of  greater 
heaviness  in  front  of  the  plate  than  would  be  computed  for 


816  MECHANICS   OF   ENGINEERING. 

the  given  temperature  and  barometric  height  for  use  in  eqs. 
(1)  and  (2) 

The  experiments  of  Borda  in  1763  led  to  the  formula 

P=  [0.0031  +  0.00035tf]#ya      ....     (3) 

for  the  total  pressure  upon  a  plate  moving  through  the  air 
in  a  direction  l  to  its  own  surface.  P  is  the  pressure  in 
pounds,  c  the  length  of  the  contour  of  the  plate  in  feet,  and  S 
its  surface  in  square,  feet,  while  v  is  the  velocity  in  miles  per 
hour.  Adopting  the  same  form  of  formula,  Hagen  found, 
from  experiments  in  18T3,  the  relation 

P=  [0.002894  -t-0.00014£]xSV       ...     (4) 

for  the  same  case  of  fluid  resistance. 

Hagen's  experiments  were  conducted  with  great  care,  but 
like  Borda's  were  made  with  a  "whirling  machine,"  in  which 
the  plate  was  caused  to  revolve  in  a  horizontal  circle  of  only 
7  or  8  feet  radius  at  the  end  of  a.  horizontal  bar  rotating  about 
a  vertical  axis.  Hagen's  plates  ranged  from  4  to  40  sq.  in.  in 
area,  and  the  velocities  from  1  to  4  miles  per  hour. 

The  last  result  was  quite  closely  confirmed  by  Mr.  H.  Allen 
Hazen  at  Washington  in  November  1886,  the  experiments 
being  made  with  a  whirling  machine  and  plates  of  from  16  to 
576  sq.  in.  area.  (See  the  American  Journal  of  Science,  Oct. 
1887,  p.  245.) 

In  Thibault's  experiments  plates  of  areas  1.16  and  1.531  sq. 
ft.  were  exposed  to  direct  wind-pressure,  giving  the  formula 

P  =  0.00475#y3 (5) 

Recent  experiments  in  France  (see  R.  R.  and  Eng.  Journal, 
Feb.  J87),  where  flat  boards  were  hung  from  the  side  of  a  rail- 
way train  run  at  different  velocities,  gave  the  formula 

P  =  0.00535/V (6) 

The  highest  velocity  was  44  miles  per  hour.  The  magnitude 
of  the  area  did  not  seemingly  affect  the  relation  given.  More 


PLATES   IN   FLUIDS.  817 

extended  and  elaborate  experiments  are  needed  in  this  field, 
those  involving  a  motion  of  translation  being  considered  the 
better,  rather  than  with  whirling  machines,  in  which  "centrif- 
ugal action"  must  have  a  disturbing  influence. 

The  notation  and  units  for  eqs.  (4),  (5),  and  (6)  are  the  same 
as  those  given  for  (3). 

It  may  be  of  interest  to  note  that  if  equation  (3)  of  §  568  be 
considered  applicable  to  this  case  of  the  pressure  of.  an  un- 
limited stream  of  fluid  against  a  plate  placed  at  right-angles  to 
the  current,  with  T^put  equal  to  the  area  of  the  plate,  we  ob- 
tain, after  reduction  to  the  units  prescribed  above  for  the  pre- 
ceding equations  and  putting  a  =  90°, 

P  =  0.0053$y2  .......     (T) 

The  value  y  =  0.0807  Ibs.  per  cub.  ft.  has  been  used  in  the 
substitution,  corresponding  to  a  temperature  of  freezing  and 
a  barometric  height  of  30  inches.  At  higher  temperatures, 
of  course,  y  would  be  less,  unless  with  very  high  barometer. 

569a.  Example.  —  Supposing  each  blade  of  the  paddle-wheel 
of  a  steamer  to  have  an  area  of  6  sq.  ft.,  and  that  when  in  the 
lowest  position  its  velocity  [relatively  to  the  water,  not  to  the 
vessel]  is  5  ft.  per  second  ;  what  resistance  is  it  overcoming  in 
salt  water  ? 

From  eq.  (1)  of  §  569,  with  C  =  1.13  and  y  =  64  Ibs.  per 
cubic  foot,  we  have  (ft.,  lb.,  sec.) 


If  on  the  average  there  may  be  considered  to  be  three  pad- 
dles always  overcoming  this  resistance  on  each  side  of  the 
boat,  then  the  work  lost  (work  of  "  slip"}  in  overcoming  these 
resistances  per  second  (i.e.,  power  lost)  is 

Z,  =  [6  X  169.4]  Ibs.  X  5  ft.  per  sec.  =  5082  ft.-lbs.  per  sec. 
or  9.24  Horse  Power  (since  5082  -5-  550  =  9.24). 


818 


MECHANICS   OF   ENGINEERING. 


If,  further,  the  velocity  of  the  boat  is  uniform  and  =  20  ft 
per  sec.,  the  resistance  of  the  water  to  the  progress  of  the  boat 
at  this  speed  being  6  X  169.4,  i.e.  1016.4  Ibs.,  the  power  ex- 
pended in  actual  propulsion  is 

Z2  =  1016.4  X  20  =  20328  ft.-lbs.  per  sec. 

Hence  the  power  expended  in  both  ways  (usefully  in  propul- 
sion, uselessly  in  "slip")  is 


t  =  25410  ft-lbs.  per  sec.  =  46.2  H.  P. 


Of  this,  9.24  H.  P.,  or  about  20  per  cent,  is  lost  in  "  slip." 

570.  Wind-pressure 
on  the  surface  of  a 
roof  inclined  at  an 
angle  =  a  with  the 
horizontal,  i.e.,  with 
the  direction  of  the 
wind,  is  usually  esti- 
mated according  to 
the  empirical  formula 


FIG.  640. 


(Button's) 


p—p'  [sin  a]  [1-84  cos  a  - 1]? 


(1) 


in  which  p'  =  pressure  of  wind  per  unit  area  against  a  vertical 
surface  (  ~|  to  wind),  and  p  —  that  against  the  inclined  plane 
(and  normal  to  it)  at  the  same  velocity.  For  a  value  of 
p'  =  40  Ibs.  per  square  foot  (as  a  maximum),  we  have  the 
following  values  for  JP,  computed  from  (1) : 


For  a  =        5° 

10° 

15° 

20° 

25° 

30° 

35° 

40° 

45° 

50° 

55° 

60° 

p=(lbs.  sq.  ft.)  5.  2 

9.6 

14 

18.3 

22.5 

26.5 

30.1 

33.4 

36.1 

38.1 

39.6 

40. 

Duchemin's  formula  for  the  normal  pressure  per  unit-area  is 

,      2  sin2  a 
p=p  •?       .  2    , (2) 


WIND   AND   SAIL. 


819 


with  the  same  notation  as  above.  Some  experimenters  in 
London  tested  this  latter  formula  by  measuring  the  pressure 
on  a  metal  plate  supported  in  front  of  the  blast-pipe  of  a  blow- 
ing engine ;  the  results  were  as  follows : 


a  = 

15° 

20° 

60° 

90° 

p  by  experiment  =  (in  Ibs.  per  sq.  ft.) 

1.65 
1.60 

2.05 
2.02 

3.01 
3.27 

3.31 
3.31 

By  Duchemin's  formula  p  = 

The  scale  of  the  Smithsonian  Institution  at  Washington  for 
the  estimation  and  description  of  the  velocity  and  pressure  of 
the  wind  is  as  follows : 


Grade. 

Velocity  iu 
miles  per  hour. 

Pressure  per 
sq.  foot  in  Ibs. 

Name. 

0 

0 

0.00 

Calm. 

1 
2 

2 
4 

0.02 

0.08 

Very  light  breeze. 
Gentle  breeze. 

3 

12 

0.75 

Fresh  wind. 

4 

25 

3.00 

Strong  wind. 

5 

35 

6 

High  wind. 

6 

45 

10 

Gale. 

7 

60 

18 

Strong  gale. 

8 

73 

Violent  gale. 

9 

90 

Hurricane. 

10 

100 

Most  violent  hurricane. 

571.  Mechanics  of  the  Sail-boat. — The  action  of  the  wind  on  a 
sail  will  be  understood  from  the  following.  Let  Fig.  641 
represent  the  boat  in  horizontal  projection  and  OS  the  sail,  O 
being  the  mast.  For  , 

simplicity  we  consider 
the  sail  to  be  a 
and  to  remain 
At  this  instant  the  boat 
is  moving  in  the  direc- 
tion MB  of  its  fore-and- 
aft  line  with  a  velocity 
=  c,  the  wind  having  a  velocity  of  the  direction  and  magni- 
tude represented  by  w  (purposely  taken  at  an  angle  <  90°  with 
the  direction  of  motion  of  the  boat).  We  are  now  to  inquire 
the  nature  of  the  action  of  the  wind  on  the  boat,  and  whether 


vertical.  ..i  \p  ot 


FIG.  641. 


820  MECHANICS   OF   ENGINEERING. 

in  the  present  petition  its  tendency  is  to  accelerate,  or  retard, 
the  motion  of  the  boat.  If  we  form  a  parallelogram  of  which 
w  is  the  diagonal  and  c  one  side,  then  the  other  -side  OK,  mak- 
ing some  angle  a  with  BM,  will  be  the  velocity  v  of  the  wind 
relatively  to  the  boat  (and  sail),  and  upon  this  (and  not  upon  w) 
depends  the  action  on  the  sail.  The  sail,  being  so  placed  that 
the  angle  0  is  smaller  than  a,  will  experience  pressure  from 
the  wind  ;  that  is,  from  the  impact  of  the  particles  of  air  which 
strike  the  surface  and  glance  along  it.  This  pressure,  P,  is 
normal  to  the  sail  (considered  smooth),  and  evidently,  for  the 
position  of  the  parts  in  the  figure,  the  component  of  P  along 
MB  points  in  the  same  direction  as  <?,  and  hence  if  that  com- 
ponent is  greater  than  the  water-resistance  to  the  boat  at  this 
velocity,  c  will  be  accelerated;  if  less,  c  will  be  retarded. 
Any  change  in  c,  of  course,  gives  a  different  form  to  the 
parallelogram  of  velocities,  and  thus  the  relative  velocity  v 
and  the  pressure  P,  for  a  given  position  of  the  sail,  will  both 
change.  [The  component  of  P  ~]  to  MB  tends,  of  course,  to 
cause  the  boat  to  move  laterally,  but  the  great  resistance  to 
such  movement  at  even  a  very  slight  lateral  velocity  will  make 
the  resulting  motion  insignificant.] 

As  c  increases,  a  diminishes,  for  a  given  amount  and  position 
of  w ;  and  the  sail  must  be  drawn  nearer  to  the  line  MB,  i.e. 
6  must  be  made  to  decrease,  to  derive  a  wind-pressure  having 
ft  forward  fore-and-aft  component ;  and  that  component  be- 
comes smaller  and  smaller.  But  if  the  craft  is  an  ice-boat,  this 
small  component  may  still  be  of  sufficient  magnitude  to  exceed 
the  resistance  and  continue  the  acceleration  of  c  until  c  is 
larger  than  w ;  i.e.,  the  boat  may  be  caused  to  go  as  fast  as,  or 
faster  than,  the  wind,  and  still  be  receiving  from  the  latter  a 
forward  pressure  which  exceeds  the  resistance.  And  it  is 
plain  that  there  is  nothing  in  the  geometry  of  the  figure  to 
preclude  such  a  relation  (i.e.,  c  >  w,  with  6  <  a  and  >  0). 

572.  Resistance  of  Still  Water  to  Moving  Bodies,  Completely 
Immersed. — This  resistance  depends  on  the  shape,  position,  and 
velocity  of  the  moving  body,  and  also  upon  the  roughness  of 
its  surface.  If  it  is  pointed  at  both  ends  (Fig.  642)  with  its 


MOVING   BODIES,    IMMERSED. 


821 


FIG.  642. 


axis  parallel  to  the  velocity,  v,  of  its  uniform  motion,  the 
stream  lines     on     closing    tor 
gether  smoothly  at  the  hinder 
extremity,  or  stern,  B^  exert 
normal   pressures  against  the 
surface  of  the  portion  CD...B 
whose     longitudinal     compo- 
nents   approximately   balance 
the  corresponding  components 
of  the  normal  pressures  on  CD  .  .  .  A  ;  so  that  the  resistance 
R,  which  must  be  overcome  to  maintain  the  uniform  velocity 
^,  is  mainly  due  to  the  "  skin-friction"  alone,  distributed  along 
the  external  surface  of  the  body ;  the  resultant  of  these  resist- 
ances is  a  force  R  acting  in  the  line  AB  of  symmetry  (sup- 
posing the  body  symmetrical  about  the  direction  of  motion). 
If,  however,  Fig.  643,  the  stern,   E..B..F\*  too   bluff, 

. eddies  are  formed  round  the  corners 

^and  Fj  and  the  pressure  on  the 
surface  E .  .  .  F  is  much  less  than 
in  Fig.  642;  i.e.,  the  water  pres- 
sure from  behind  is  less  than  the 
backward  (longitudinal)  pressures 
from  in  front,  and  thus  the  resultant 
resistance  R  is  due  partly  to  skin- 
friction  and  partly  to  "  eddy-making"  (§  569). 

[NOTE. — The  diminished  pressure  on  EF\&  analogous  to  the 
loss  of  pressure  of  water  (flowing  in  a  pipe)  after  passing  a  nar- 
row section  the  enlargement 
from  which  to  the  original 
section  is  sudden.  E.g.,  Fig. 
644,  supposing  the  velocity  v 
and  pressure  p  (per  unit-area) 
to  be  the  same  respectively 
at  A  and  A\  in  the  two 
pipes  shown,  with  diameter 
AL  =  WK=AfLf  =  WK' ; 
then  the  pressure  at  M  is  FrG.  644. 

equal  to  that  at  A  (disregarding  skin-friction),  whereas  that  at 


FIG.  643. 


822 


MECHANICS   OF  ENGINEERING. 


' 


M '  is  considerably  less  than  that  at  Af  on  account  of  the  head 
lost  in  the  sudden  enlargement. .  (See  also  Fig.  575.)] 

It  is  therefore  evident  that  Ihiffness  of  stern  increases  the 
resistance  much  more  than  Huff  ness  of  bow. 

In  any  case  experiment  shows  that  for  a  given  body  sym- 
metrical about  an  axis  and  moving  through  a  fluid  (not  only 
water,  but  any  fluid)  in  the  direction  of  its  axis  with  a  uni- 
form velocity  =  v,  we  may  write  approximately  the  resistance 


=  (resistance  at  vel.  v)  = 


As  in  preceding  paragraphs,  F  =  area  of  the  greatest  section, 
"I  to  axis,  of  the  external  surface  of  body  (not  of  the  sub- 
stance) ;  i.e.,  the  sectional  area  of  the  circumscribing  cylinder 
(cylinder  in  the  most  general  sense)  with  elements  parallel  to 
the  axis  of  the  body,  y  ==.  the  heaviness  (§  409)  of  the  fluid, 
and  v  =  velocity  of  motion ;  while  £  is  an  abstract  number 
dependent  on  experiment. 

According  to  Weisbach,  who  cites  different  experimenters, 
we  can  put  for  spheres,  moving  in  water,  £  =  about  0.55 ;  for 
cannon-balls  moving  in  water,  £  =  .467. 

According  to  Robins  and  Hutton,  for  spheres  in  air,  we 
have 


For  0  in  mets.  )  .. 
per  sec.        j" 

5 

25 

100 

200 

300 

400 

rnn  j  metres 
U  1  per  sec. 

C=.59 

.63 

.67 

.71 

.77 

.88 

.99 

1.04 

For  musket-balls  in  the  air,  Piobert  found 

£  =  0.451  (1  4-  0.0023  x  veloc.  in  metres  per  sec.). 
From  Dubuat's  experiments,  for  the  resistance  of  water  to  a 
right  prism  moving  endwise  and  of  length  =  Z, 


For  (I:  V'f)=   0 

1 

2 

3 

5  =  1.25 

1.26 

1.31 

1.33 

For  a  circular  cylinder  moving  perpendicularly  to  its  axis 
Borda  claimed  that  £  is  one-half  as  much  as  for  the  circura- 


MOVING  BODIES,    IMMERSED.  823 

scribing  right  parallelepiped  moving  with  four  faces  parallel 
to  direction  of  motion. 

EXAMPLE.  —  The  resistance  of  the  air  at  a  temperature  of 
freezing  and  tension  of  one  atmosphere  to  a  musket-ball  \  inch 
in  diameter  when  moving  with  a  velocity  of  328  ft.  per  sec. 
is  thus  determined  by  Piobert's  formula,  above  : 

C  =  0.451(1  +  .0023  X  100)  =  0.554  ; 
hence,  from  eq.  (1), 

R  =  0.554  X  X  -0807  X  =  al018  lbs' 


572a.  Deviation  of  a  Spinning  Ball  from  a  Vertical  Plane  in 
Still  Air.  —  It  is  a  well-known  fact  in  base-ball  playing  that  if  a 
rapid  spinning  motion  is  given  to  the  ball  about  a  vertical  axis 
as  well  as  a  forward  motion  of  translation,  its  path  will  not 
remain  in  its  initial  vertical  plane,  but  curve  out  of  that  plane 
toward  the  side  on  which  the  absolute  velocity  of  an  external 
point  of  the  ball's  surface  is  least.  Thus,  if  the  ball  is  thrown 
from  North  to  South,  with  a  spin  of  such  character  as  to  ap- 
pear "  clock-wise"  seen  from  above,  the  ball  will  curve  toward 
the  West,  out  of  the  vertical  plane  in  which  it  started. 

This  could  not  occur  if  the  surface  of  the  ball  were  perfectly 
smooth  (there  being  also  no  adhesion  between  that  surface  and 
the  air  particles),  and  is  due  to  the  fact  that  the  cushion  of  com- 
pressed air  which  the  ball  piles  up  in  front  during  its  progress, 
and  which  would  occupy  a  symmetrical  position  with  respect 
to  the  direction  of  motion  of  the  centre  of  the  ball  if  there 
were  no  motion  of  rotation  of  the  kind  indicated,  is  now  piled 
up  somewhat  on  the  East  of  the  centre  (in  example  above), 
creating  constantly  more  obstruction  on  that  side  than  on  the 
right  ;  the  cause  of  this  is  that  the  absolute  velocity  of  the  sur- 
face-points, at  the  same  level  as  the  centre  of  ball,  is  greatest, 
and  the  friction  greatest,  at  the  instant  when  they  are  passing 
through  their  extreme  Easterly  positions;  since  then  that 
velocity  is  the  sum  of  the  linear  velocity  of  translation  and 
that  of  rotation  ;  whereas,  in  the  position  diametrically  oppo- 


824  MECHANICS    OF   ENGINEERING. 

site,  on  the  West  side,  the  absolute  velocity  is  the  difference  ; 
hence  the  greater  accumulation  of  compressed  air  on  the  left 
(in  the  case  above  imagined,  ball  thrown  from  North  to  South, 

etc.). 

573.  Robinson's  Cup-anemometer. — This  instrument,  named 
after  Dr.  T.  R.  Robinson  of  Armagh,  Ireland,  consists  of  four 
hemispherical  cups  set  at  equal  intervals  in  a  circle,  all  facing 
in  the  same  direction  round  the  circle,  and  so  mounted  on  a 
light  but  rigid  framework  as  to  be  capable  of  rotating  with 
but  little  friction  about  a  vertical  axis.  When  in  a  current  of 
air  (or  other  fluid)  the  apparatus  begins  to  rotate  with  an  ac- 
celerated velocity  on  account  of  the  pressure  against  the  open 
mouth  of  a  cup  on  one  side  being  greater  than  the  resistance 
met  by  the  back  of  the  cup  diametrically  opposite.  Yery  soon,, 
however,  the  motion  becomes  practically  uniform,  the  cnp- 
centre  having  a  constant  linear  velocity  v"  the  ratio  of  which 
to  the  velocity,  v',  of  the  wind  at  the  same  instant  must  be 
found  in  some  way,  in  order  to  deduce  the  value  of  the  latter 
from  the  observed  amount  of  the  former  in  the  practical  use 
of  the  instrument.  After  sixteen  experiments  made  by  Dr. 
Robinson  on  stationary  cups  exposed  to  winds  of  varying  in- 
tensities, from  a  gentle  breeze  to  a  hard  gale,  the  conclusion 
was  reached  by  him  that  with  a  given  wind- velocity  the  total 
pressure  on  a  cup  with  concave  surface  presented  to  the  wind 
was  very  nearly  four  times  as  great  as  that  exerted  when  the 
convex  side  was  presented,  whatever  the  velocity  (see  vol. 
xxu  of  Transac.  Irish  Royal  Acad.,  Pa?^t  /,  p.  163). 

Assuming  this  ratio  to  be  exactly  4.0  and  neglecting  axle- 
friction,  we  have  the  data  for  obtaining  an  approximate  value 
of  m,  the  ratio  of  vf  to  the  observed  v",  when  the  instrument  is 
in  use.  The  influence  of  the  wind  on  those  cups  the  planes  of 
whose  mouths  are  for  the  instant  ||  to  its  direction  will  also  be 
neglected. 

If,  then,  Fig.  645,  we  write  the  impulse  on  a  cup  when  the 
hollow  is  presented  to  the  wind  [§  572,  eq.  (1)] 

'" 


ROBINSON'S  CUP-ANEMOMETER. 
and  the  resistance  when  the  convex  side  is  presented 


825 


we  may  also  put 


(3) 


In  (1)  and  (2)  v,  and  v9  are  relative  velocities. 

^Regarding  only  the  two  cups  A  and  B, 
whose  centres  at  a  definite  instant  are  mov- 
ing in  lines  parallel  to  the  direction  of  the 
wind,  it  is  evident  that  the  motion  of  the 

cups  does  not  become  uniform  until  the  rel- >v 

ative  velocity  v'  —  v"  of  the  wind  and  cup 
A  (retreating  before  the  wind)  has  become 
so  small,  and  the  relative  velocity  v'  4-  v" 
with  which  B  advances  to  meet  the  air- 
particles  has  become  so  great,  that  the  im- 
pulse of  the  wind  on  A  equals  the  resist- 
ance encountered  by  B;  i.e.,  these  forces, 
Ph  and  Pc ,  must  be  equal,  having  equal 
lever-arms  about  the  axis.  Hence,  for  uniform  rotary  motion, 


i.e.  [see  eq.  (3)], 

4     ~  *  '= 


or 


Solving  the  quadratic  for  m,  we  obtain 

TW  =  3.00  ........     (6) 

That  is,  the  velocity  of  the  wind  is  about  three  times  that  of 
the  cup-centre. 

574.  Experiments  with  Robinson's  Cup-anemometer.  —  The 
ratio  3.00  just  obtained  is  the  one  in  common  use  in  connec- 
tion with  this  instrument  in  America.  Experiments  by  Mr. 


826  MECHANICS   OF   ENGINEERING. 

Hazen  at  Washington  in  1886  (Am.  Jour.  Science,  Oct.  '87, 
p.  248)  were  made  on  a  special  type  devised  by  Lieut.  Gibbon. 
The  anemometer  was  mounted  on  a  whirling  machine  at  the 
end  of  a  16-ft.  horizontal  arm,  and  values  for  m  obtained,  with 
velocities  up  to  12  miles  per  hour,  from  2.84  to  3.06 ;  average 
2.94.  The  cups  were  4  in.  in  diameter  and  the  distance  of  their 
centres  from  the  axis  6.72  in.,  these  dimensions  being  those 
usually  adopted  in  America.  This  instrument  was  nearly  new 
and  was  well  lubricated. 

Dr.  Robinson  himself  made  an  extensive  series  of  experi- 
ments, with  instruments  of  various  sizes,  of  which  an  account 
may  be  found  in  the  Philos.  Transac.  for  1878,  p.  797  (see 
also  the  volume  for  1880,  p.  1055).  Cups  of  4  in.  and  also  of 
9  in.  were  employed,  placed  first  at  24  and  then  at  12  in.  from 
the  axis.  The  cup-centres  revolved  in  a  (moving)  vertical 
plane  perpendicular  to  the  horizontal  arm  of  a  whirling- 
machine  ;  this  arm,  however,  was  only  9  ft.  long.  A  friction- 
brake  was  attached  to  the  axis  of  the  instrument  for  testing  the 
effect  of  increased  friction  on  the  value  of  in.  At  high  speeds 
of  30  to  40  miles  per  hour  (i.e.,  the  speed  of  the  centre  of  the 
instrument  in  its  horizontal  circle,  representing  an  equal  speed 
of  wind  for  an  instrument  in  actual  use  with  axis  stationary) 
the  effect  of  friction  was  relatively  less  than  at  low  velocities. 
That  is,  at  high  speeds  with  considerable  friction  the  value  of 
m  was  nearly  the  same  as  with  little  friction  at  low  speeds. 
With  the  large  9  in.  cups  at  a  distance  of  either  24  or  12  in. 
from  the  axis  the  value  of  m  at  30  miles  per  hour  ranged 
generally  from  2.3  to  2.6,  with  little  or  much  friction  ;  while 
with  the  minimum  friction  m  rose  slowly  to  about  2.9  as  the 
velocity  diminished  to  10  miles  per  hour.  At  5  miles  per 
hour  with  minimum  friction  m  was  3.5  for  the  24  in.  instru- 
ment and  about  5.0  for  the  12  in.  The  effect  of  considerable 
friction  at  low  speeds  was  to  increase  m,  making  it  as  high  as 
8  or  10  in  some  cases.  With  the  4  in.-cups  no  value  was  ob- 
tained for  m  less  than  3.3.  On  the  whole,  Dr.  Robinson  con- 
cluded that  m  is  more  likely  to  have  a  constant  value  at  all 
velocities  the  larger  the  cups,  the  longer  the  arms,  and  the  less 
the  friction,  of  the  anemometer.  But  few  straight-line  experi- 


VARIOUS   ANEMOMETERS.  827 

ments  have  been  made  with  the  cup-anemometer,  the  most 
noteworthy  being  mentioned  on  p.  308  of  the  Engineering 
News  for  October  188T.  The  instrument  was  placed  on  the 
front  of  the  locomotive  of  a  train  running  between-  Baltimore 
and  Washington  on  a  calm  day.  The  actual  distance  is  40 
miles  between  the  two  cities,  while  from  the  indications  of  the 
anemometer,  assuming  m  =  3.00,  it  would  have  been  in  one  trip 
46  miles  and  in  another  47.  The  velocity  of  the  train  was  20 
miles  per  hour  in  one  case  and  40  in  the  other. 

575  Other  Anemometers. — Both  Biram's  and  Castello's  ane- 
mometers consist  of  a  wheel  furnished  with  radiating  vanes 
set  obliquely  to  the  axis  of  the  wheel,  forming  a  small  "wind- 
mill," somewhat  resembling  the  current-meter  for  water  shown 
in  Fig.  604 ;  having  six  or  eight  blades,  however.  The  wheel 
revolves  with  but  little  friction,  and  is  held  in  the  current  of 
air  with  its  axis  parallel  to  the  direction  of  the  latter,  and  very 
quickly  assumes  a  steady  motion  of  rotation.  The  number  of 
revolutions  in  an  observed  time  is  read  from  a  dial.  The  in- 
struments must  be  rated  by  experiment,  and  are  used  chiefly 
in  measuring  the  velocity  of  the  currents  of  air  in  the  galleries 
of  mines,  of  draughts  of  air  in  flues  and  ventilating  shafts,  etc. 

To  quote  from  vol.  v  of  the  Keport  of  the  Geological  Sur- 
vey of  Ohio,  p.  370:  "Approximate  measurements  (of  the 
velocity  of  air)  are  made  by  miners  by  flashing  gunpowder, 
and  noting  with  a  watch  the  speed  with  which  the  smoke 
moves  along  the  air-way  of  the  mine.  A  lighted  lamp  is 
sometimes  used,  the  miner  moving  along  the  air-gallery,  and 
keeping  the  light  in  a  perfectly  perpendicular  position,  noting 
the  time  required  to  pass  to  a  given  point." 

Another  kind  makes  use  of  the  principle  of  Pitot's  Tube 
(p.  751),  and  consists  of  a  U -tube  partially  filled  with  water, 
one  end  of  the  tube  being  vertical  and  open,  while  the  other 
turns  horizontally,  and  is  enlarged  into  a  wide  funnel,  whose 
mouth  receives  the  impulse  of  the  current  of  air ;  the  differ- 
ence of  level  of  the  water  in  the  two  parts  of  the  U  is  a  meas- 
ure of  the  velocity. 


828  MECHANICS   OF  ENGINEERING. 

576.  Resistance  of  Ships. — We  shall  first  suppose  the  ship  to 
be  towed  at  a  uniform  speed ;  i.e.,  to  be  without  means  of  self- 
propulsion  (under  water).  This  being  the  case,  it  is  found  that 
at  moderate  velocities  (under  six  miles  per  hour),  the  ship 
being  of  "fair"  form  (i.e.,  the  hull  tapering  both  at  bow  and 
stern,  under  water)  the  resistance  in  still  water  is  almost  wholly 
due  to  skin-friction,  "eddy-making"  (see  §  569)  being  done 
away  with  largely  by  avoiding  a  bluff  stern. 

When  the  velocity  is  greater  than  about  six  miles  an  hour 
the  resistance  is  much  larger  than  would  be  accounted  for  by 
skin-friction  alone,  and  is  found  to  be  connected  with  the  sur- 
face-disturbance or  waves  produced  by  the  motion  of  the  hull 
in  (originally)  still  water.  The  recent  experiments  of  Mr. 
Froude  and  his  son  at  Torquay,  England,  with  models,  in  a  tank 
300  feet  long,  have  led  to  important  rules  (see  Mr.  White's 
Naval  Architecture  and  "Hydromechanics"  in  the  Ency. 
Britann.}  of  so  proportioning  not  only  the  total  length  of  a 
ship  of  given  displacement,  but  the  length  of  the  entrance  (for- 
ward tapering  part  of  hull)  and  length  of  run  (hinder  tapering 
part  of  hull),  as  to  secure  a  minimum  "wave-making  resist- 
ance" as  this  source  of  resistance  is  called. 

To  quote  from  Mr.  White  (p.  460  of  his  Naval  Architecture, 
London,  1882):  "Summing  up  the  foregoing  remarks,  it 
appears : 

"  (1)  That / rictional  resistance,  depending  upon  the  area  of 
the  immersed  surface  of  a  ship,  its  degree  of  roughness,  its 
length,  and  (about)  the  square  of  its  speed,  is  not  sensibly 
affected  by  the  forms  and  proportions  of  ships ;  unless  there 
be  some  unwonted  singularity  of  form,  or  want  of  fairness. 
For  moderate  speeds  this  element  of  resistance  is  by  far  the 
most  important ;  for  high  speeds  it  also  occupies  an  important 
position — from  50  to  60  per  cent  of  the  whole  resistance, 
probably,  in  a  very  large  number  of  classes,  when  the  bottoms 
are  clean  •  and  a  larger  percentage  when  the  bottoms  become 
foul. 

"  (2)  That  eddy-maldng  resistance  is  usually  small,  except  in 
special  cases,  and  amounts  to  8  or  10  per  cent  of  the  frictional 


RESISTANCE   OF  SHIPS.  829 

resistance.  A  defective  form  of  stern  causes  largely  increased 
eddy-making. 

"  (3)  That  wave-making  resistance  is  the  element  of  the 
total  resistance  which  is  most  influenced  by  the  forms  and  pro- 
portions of  ships.  Its  ratio  to  the  frictional  resistance,  as  well 
as  its  absolute  magnitude,  depend  on  many  circumstances ;  the 
most  important  being  the  forms  and  lengths  of  the  entrance 
and  run,  in  relation  to  the  intended  full  speed  of  the  ship. 
For  every  ship  there  is  a  limit  of  speed  beyond  which  each 
small  increase  in  speed  is  attended  by  a  disproportionate  in- 
crease in  resistance ;  and  this  limit  is  fixed  by  the  lengths  of 
the  entrance  and  run — the  4  wave-making  features  '  of  a  ship. 

"  The  sum  of  these  three  elements  constitutes  the  total  re- 
sistance offered  by  the  water  to  the  motion  of  a  ship  towed 
through  it,  or  propelled  by  sails ;  in  a  steamship  there  is  an 
'  augment '  of  resistance  due  to  the  action  of  the  propel- 
lers." 

In  the  case  of  a  steamship  driven  by  a  screw  propeller,  this 
augment  to  the  resistance  varies  from  20  to  45  per  cent  of  the 
"  tow-rope  resistance,"  on  account  of  the  presence  and  action 
of  the  propeller  itself ;  since  its  action  relieves  the  stern  of 
some  of  the  forward  hydrostatic  pressure  of  the  water  closing 
in  around  it.  Still,  if  the  screw  is  placed  far  back  of  the  stern, 
the  augment  is  very  much  diminished  ;  but  such  a  position  in- 
volves risks  of  various  kinds  and  is  rarely  adopted. 

We  may  compute  approximately  the  resistance  of  the  water 
to  a  ship  propelled  by  steam  at  a  uniform  velocity  -y,  in  the 
following  manner :  Let  L  denote  the  power  developed  in  the 
engine  cylinder ;  whence,  allowing  10  per  cent  of  L  for  engine 
friction,  and  15  per  cent  for  "  work  of  slip"  of  the  propeller- 
blade,  we  have  remaining  0.75Z,  as  expended  in  overcoming 
the  resistance  R  through  a  distance  =  v  each  unit  of  time ;  i.e., 

(approx.)  0.75Z  =  7fo (1) 

EXAMPLE.— If  3000  indicated  H.  P.  (§  132)  is  exerted  by  the 
engines  of  a  steamer  at  a  uniform  speed  of  15  miles  per  hour 


330  MECHANICS    OF    ENGINEERING. 

(=  22  ft.  per  sec.),  we  have  (with  above  allowances  for  slip  and 
engine  friction)  [foot-lb.-sec.] 

|  X  3000  X  550  =  E  X  22 ;     .-.    R  =  56250  Ibs. 

Further,  since  J2  varies  (roughly)  as  the  square  of  the  veloc- 
ity, and  can  therefore  be  written  It  =  (Const.)  X  'y2,  we  have 
from  (1) 

L  =  a  constant  X  vs (2) 

as  a  roughly  approximate  relation  between  the  speed  and  the 
power  necessary  to  maintain  it  uniformly.  In  view  of  eq.  (3) 
involving  the  cube  of  the  velocity  as  it  does,  we  can  understand 
why  a  large  increase  of  power  is  necessary  to  secure  a  propor- 
tionally small  increase  of  speed. 

577.  "  Transporting  Power,"  or  Scouring  Action,  of  a  Current. 
— The  capacity  or  power  of  a  current  of  water  in  an  open 
channel  to  carry  along  with  it  loose  particles,  sand,  gravel, 
pebbles,  etc.,  lying  upon  its  bed  was  investigated  experimen- 
tally by  Dubuat  about  a  century  ago,  though  on  a  rather  small 
scale.  His  results  are  as  follows : 

The  velocity  of  current  must  be  at  least 

0.25  ft.  per  sec.,  to  transport  silt ; 

0.50        «  "  «         loam; 

1.00        "  "          «        sand; 

2.00        "  "          «        gravel; 

3.5          «  "          «         pebbles  1  in.  in  diam.; 

4.0          "  "  "         broken  stone : 

5.0          «  "          «        chalk,  soft  shale. 

However,  more  modern  writers  call  attention  to  the  fact 
that  in  some  instances  beds  of  sand  are  left  undisturbed  by 
currents  of  greater  velocity  than  that  above  indicated  for  sand, 
and  explain  this  fact  on  the  theory  that  the  water-particles 
may  not  move  parallel  to  the  bed,  but  in  cycloids,  approxi- 
mately, like  the  points  in  the  rim  of  a  rolling  wheel,  so  as  to 
have  little  or  no  scouring  action  on  the  bed  in  those  cases. 

In  case  the  particles  move  in  filaments  or  stream-lines 
parallel  to  the  axis  of  the  stream  the  statement  is  sometimes 
made  that  the  "  transporting  power"  varies  as  the  sixth  power 


"  TRANSPORTING  POWER"   OF   CURRENTS. 


831 


FIG.  646. 


of  the  velocity  of  the  current,  by  which  is  meant,  more  defi- 
nitely, the  following :  Fig.  646.  Conceive  a  row  of  cubes  (or 
other  solids  geometri- 
cally similar  to  each 
other)  of  many  sizes, 
all  of  the  same  heavi- 
ness (§  7),  and  simi- 
larly situated,  to  be 
placed  on  the  horizon- 
tal bottom  of  a  trough 
and  there  exposed  to 
a  current  of  water,  *** 
being  completely  im- 
mersed. Suppose  the  coefficient  of  friction  between  the  cubes 
and  the  trough-bottom  to  be  the  same  for  all.  Then,  as  the 
current  is  given  greater  and  greater  velocity  v,  the  impulse 
Pm  (corresponding  to  a  particular  velocity  vm)  against  some 
one,  m,  of  the  cubes,  will  be  just  sufficient  to  move  it,  and  at 
some  higher  velocity  vn  the  impulse  Pn  against  some  larger 
cube,  n,  will  be  just  sufficient  to  move  it,  in  turn.  "We  are  to 
prove  that  Pm  :  Pn  ::  vm6  :  vn\ 

Since,  when  a  cube  barely  begins  to  move,  the  impulse  is 
equal  to  the  friction  on  its  base,  and  the  frictions  under  the 
cubes  (when  motion  is  impending)  are  proportional  to  their 
volumes  (see  above),  we  have  therefore 


(i) 


Also,  the  impulses  on  the  cubes,  whatever  the  velocity,  are  pro- 
portional to  the  face  areas  and  to  the  squares  of  the  velocities 
(nearly  ;  see  §  572) ;  hence 


Pn          ft.X' 


....    (2) 


From  (1)  and  (2)  we  have 


=  ^;   •    •    •   •    (3) 


'w  "** 


832  MECHANICS   OF   ENGINEERING. 

while  from  (3)  and  (2)  we  have,  finally, 

Pm:Pn::Vjn6:^c (4) 

Thus  we  see  in  a  general  way  why  it  is  that  if  the  velocity 
of  a  stream  is  doubled  its  transporting  power  is  increased 
about  sixty-four-fold  ;  i.e.,  it  can  now  impel  along  the  bottom 
pebbles  that  are  sixty-four  times  as  heavy  as  the  heaviest  which 
it  could  move  before  (of  same  shape  and  heaviness). 

Though  rocks  are  generally  from  two  to  three  times  as 
heavy  as  water,  their  loss  of  weight  under  water  causes  them  to 
encounter  less  friction  on  the  bottom  than  if  not  immersed. 


INDEX   TO    HYDRAULICS. 


Absolute  Temperature 606 

Absolute  Zero 606 

Accumulator,  Hydraulic 700 

Adiabatic  Change,  621,  629,  631,  636 
Adiabatic  Expansion  in   Com- 
pressed-air Engine 631 

Adiabatic  Flow  of  Gases  from 

an  Orifice 778 

Air  Collecting  in  Water-pipes, 

731,  736 
Air,  Compressed,  Transmission 

of 786-790 

Air-compressor 636 

Air-profile 749 

Air-pump,  Sprengel's 656 

Air-thermometer 604 

Amplitude  of  Backwater 772 

Anemometer,  Robinson's. .  824,  826 

Anemometer,  Biram's 827 

Anemometer,  Castello's 827 

Angle  of  Repose 572 

Angular  Stability  of  Ships 597 

Atkinson  Gas-engine 642,  643 

Atmosphere  as  a  Unit  Pressure,  519 
Augment  of  Resistance  of  Screw 

Propeller 829 

Backwater 771 

Ball,  Spinning,  Deviation  from 

Vertical  Plane 823 

Balloon 644 

Barker's  Mill 672 

Barometers 530 

Barometric  Levelling 619 

Baziu,  Experiments."" 688 

Beaufoy's  Experiments 814 

Bellinger,    Prof.,   Experiments 

with  Elbows 729 

Bends,  Loss  of  Head  due  to 728 

Bends  in  Open  Channels 770 

3ent  Tube,  Liquids  in 529 

Bernoulli's  Theorem   and    the 

Conservation  of  Energy 717 

Bernoulli's  Theorem  for  Gases,  773 
Bernoulli's    Theorem.    General 

Form 706 

Bernoulli's     Theorem,     Steady 

Flow  without  Friction. .  .652,  654  j 


Bernoulli's  Theorem  with  Fric- 
tion  696 

Bidone,  Experiments  on  Jets. . .  803 

Blowing-engine,  Test 776 

Borda's  Formula 722 

Bourdon  Steam-gauge 532 

Boyle's  Law , . .  615 

Bramah  Press 526 

Branching  Pipes . .  736 

Bray  ton's  Petroleum-engine  . . .  641 

Buoyant  Effort 586 

Buoyant  Effort  of  the  Atmos- 
phere   644 

Canal  Lock,  Time  of  Filling,  739,  740 

Centigrade  Scale 605 

Centre  of  Buoyancy 586 

Centre  of  Pressure 546 

Change  of  State  of  Gas 610 

Chezy's  Formula 714 

Chezy's  Formula  for  Open  Chan- 
nels   758 

Church  and  Fteley,  Report  on 
Quaker  Bridge  Darn,  558,  563,  564 

Clearance 627 

Closed  Air-manometer 614 

Coal  Consumption 643 

Coal,  Heat  of  Combustion 643 

Coefficient  of  Contraction 659 

Coefficients  of  Efflux,  661,  712, 

676,  734,  738,  784 
Coefficient  of  Fluid  Friction,  707,  797 

Coefficient  of  Resistance 704 

Coefficient  of  Roughness. .  .759,  760 
Coefficient  of  Velocity,  661,  689, 

704,  712,  723,  734 

Collapse  of  Tubes 538 

Communicating  Prismatic  Ves- 
sels   739 

Complete  and  Perfect  Contrac- 
tion   676 

Component  of  Fluid  Pressure..  525 

Compressed-air  Engine 631 

Compressed  Air,  Transmission 

of.    786 

Compressibility  of  Water 516 

Conical  Short  Tubes 692 

Conservation    of    Energy    and 


INDEX  TO   HYDRAULICS. 


PAGE 

Bernoulli's  Theorem 717 

Contraction 659 

Contraction,  Perfect 676 

Cooling  in   Sudden  Expansion 

of  Gas 622 

Critical  Temperature  of  a  Vapor,  608 

Croton  Aqueduct,  Slope 749 

Cunningham,   Experiments  on 

the  River  Ganges 759 

Cup-anemometer 824,  826 

Cups,  Impulse  of  Jet  on ...  804,  808 
Current, Transporting  Power  of,  830 

Current-meters 750 

Curved  Dams 562 

Cylinders,       Thin        Hollow, 

Strength 538 

Dams,  High  Masonry 562 

Darcy,  Experiments  with  Pitot's 

Tube 804 

Darcy  and  Bazin,  Experiments 

with  Open  Channels 758 

Decrease    of    Tension    of  Gas 

along  a  Pipe 793 

Depth  of  Flotation 592 

Deviation  of  Spinning  Ball  from 

Vertical  Plane 823 

Diaphragm  in  a  Pipe,  Loss  of 

Head 724 

Displacement  of  a  Ship 596 

Divergent  Tubes,  Flow  through,  692 
Dividing  Surface  of  Two  Fl  uids,  528 

Diving-bell 617 

Double  Floats 750 

Draught  of  Ships 596 

Dubuat's  Experiments,  707,  81 5, 

822,  830 

Duchemin's  Formula  for  Wind- 
pressure 818 

Duty  of  Pumping-engines. 644 

Dynamics  of  Gaseous  Fluids,  773-797 

Earth  Pressure 5T2 

Earthwork  Dam 566 

Eddy-making  Resistance . .  814,  828 

Efficiency 637,  642 

Efflux  of  Gases 773-797 

Efflux  from  Steam-boiler 664 

Efflux  from  Vessel  in  Motion. .  670 

Efflux  into  Condenser 665 

Efflux  under  Water 669 

Egg-shaped  Section  for  Sewers,  765 

Elastic  Fluids 516 

Elbows,  Loss  of  Head  due  to,  727,  729 
Ellis,  Book  on  Fire-streams  . . .  716 
Emptying  Vessels,  Time  of. .  737-746 

Engine,  Gas- 641 

Engine,  Hot-air 639 

Engine,  Petroleum- 641 

Engine,  Steam- 624 

Enlargement,  Sudden,  in  Pipe,  721 


Equal    Transmission    of  Fluid 

Pressure 524 

Equation  of  Continuity,  648,  737,  756 
Equation    of     Continuity     for 

Gases 773 

Equation     of    Continuity     for 

Open  Channels 756 

Equilibrium  of  Flotation 590 

Ericsson's  Hot-air  Engine 640 

Ewart's  Experiments  on  Jets .  . .  ^00 

Expanding  Steam 625 

Fahrenheit  Scale 605 

Fairbairn's  Experiments  on  Col- 
lapse of  Tubes 538 

Fanning,  Table  of  Coefficients 

of  Fluid  Friction 709 

Feet  and  Meters,  Table ...  677 

Fire-engine  Hose,  Friction  in. ..  716 

Fire-streams,  by  Ellis 71  & 

Floating  Staff 750 

Flood-gate 553 

Flotation 590 

Flow  in  Plane  Layers 648,  6o2 

Flow  in  Open  Channels 749 

Flow  of  Gas  in  Pipes 786,  790 

Fluid  Friction 695,  797,  828 

Fluid  Friction,    Coefficient  for 

Natural  Gas 797 

Fluid   Pressure,   Equal   Trans- 
mission of 524 

Force-pump 667 

Francis'  Formula  for  Overfalls,  687 
Free  Surface  of  Liquid  at  Rest,  528 

Free  Surface  a  Paraboloid 544 

Fresh  Water,  Heaviness,  Table,  518 
Friction-head  in  Open  Channels,  757 

Friction-head  in  Pipes 699 

Friction,  Fluid 695,  797,  828 

Fronde's  Experiments  on  Fluid 

Friction 696 

Fronde's  Experiments  on  Grad- 
ual Enlargement  in  Pipes  . . .  725 
Fronde's     Experiments      with 

Piezometers 720 

Fteley    and   Stearns's    Experi- 
ments on  Overfalls 687 

Fteley    and    Stearns's    Experi- 
ments with  Open  Channels. .  758 

Gas  and  Vapor 607 

Gas-engines 641 

Gaseous  Fluids 604-645 

Gas,  Flow  through  Short  Pipes,  784 
Gas,  Flow  through  Orifices.  .773-784 

Gas,  Illuminating 517,  533 

Gas,  Natural,  Flow  in  Pipes,  786,  79g 

Gas,  Steady  Flow 773 

Gas,  Velocity  of  Approach 784 

Gases,  Definition 515 

Gauging  of  Streams 755- 


INDEX  TO   HYDRAULICS. 


XI 


PAGE 

Gay-Lussac's  Law 609 

Gradual  Enlargement  in  Pipe..  725 

Granular  Materials 572 

Graphic  Representation  of 

Change  of  State  of  Gas 628 

Head  of  Water 530 

Heat-engines,  Efficiency 642 

Heaviness  of  Fluids 517 

Heaviness  of  Fresh  Water  at 

Different  Temperatures 518 

Height  Due  to  Velocity 649 

Height  of  the  Homogeneous 

Atmosphere 620 

Herschel'sVenturi Water-meter,  726 

High  Masonry  Dams 562 

Hoop-tension 537 

Hose,  Rubber  and  Leather,  Fric- 
tion in 716 

Hot-air  Engines 639 

Humphreys  and  Abbot's  Survey 

of  the  Mississippi  River... 754,  759 

Hurdy-gurdy,  California 809 

Button's  Formula  for  Wind- 
pressure  . .  818 

Hydraulic  Grade-line 715 

Hydraulic  Mean  Depth,  698,757,  764 

Hydraulic  Press 526 

Hydraulic  Radius 698 

Hydraulic  Radius  for  Minimum 

Friction 764 

Hydraulics,  Definition 518 

Hydrodynamics 646-832 

Hydrodynamics  or  Hydrokinet- 

ics.  Definition 518 

Hydromechanics,  Definition...  519 

Hydrometers 591 

Hydronietric  Pendulum 753 

Hydrostatic  Pressure 522 

Hydrostatics,  Definition 518 

Ice-making  Machine 624 

Illuminating  Gas .517,  533 

Immersion  of  Rigid  Bodies. . . .  586 

Imperfect  Contraction 680,  684 

Impulse  and  Resistance  of 

Fluids 797-832 

Impulse  of  Jet  on  Vanes 801,  805 

Inclin^i  Short  Tubes,  Efliux 

through 691 

Incomplete  Contraction 679,  684 

Inelastic  Fluids 516 

Irres'ilar  Shape,  Emptying  of 

Vesselsof 746 

Isothermal  Change 615,  629,  639 

Isothermal  Expansion 624,  635 

Isothermal  Flow  of  Gas  in 

Pipes 790 

Isothermal  Flow  of  Gases 

through  Orifices 777 

Jacket  of  Hot  Water 635 


PAGE 

Jackson's  Works  on  Hydraulics,  761 

Jet  from  Force-pump 667 

Jets,  Impulse  of 800,  803,  810 

Jets  of  Water 660,  662 

Joule,  Experiment  on  Flow  of 

Gas 782 

Kansas  City  Water-works;    Si- 
phon  731,  736 

Kinetic  Energy 672,  718 

Kinetic  Energy  of  Jet 672,  808 

Kinetic  Theory  of  Gases 516, 

606,  622 

Kutter's  Diagram 761 

Kutter's  Formula  759 

Kutter's  Hydraulic  Tables 761 

Laminated  Flow 648,  652 

Land-ties 585 

Law  of  Charles 609 

Levelling,  Barometric 619 

Liquefaction  of  Oxygen 609 

Liquid,  Definition 515 

Long    Pipes,    Flow    of    Water 

through 710-716 

Loss  of  Head 698,  703,  721,  etc. 

Loss  of  Head  Due  to  Bends 728 

Loss  of  Head  Due  to  Elbows, 

727,  729 

Loss  of  Head  Due  to  Throttle- 
valves 730 

Loss  of  Head  Due  to  Valve- 
gates 730 

Manometers 530 

Mariotte's  Law 615,  777 

Mechanics  of  the  Sail-boat 819 

Mendeleieff's  Device  for  Spec- 
ula   544 

Metacentre  of  a  Ship 599 

Metres  and  Feet,  Table 677 

Mill,  Barker's 672 

Minimum  Frictional  Resistance 

in  Open  Channel 764,  766 

Mississippi     River,     Hydraulic 

Survey 754,  759,  770 

Mixture  of  Gases 618 

Momentum,  Principle  of 812 

Moving  Pistons 524 

Moving  Vane,  Impulse  of  Jet  on  805 

Napier  on  Flow  of  Steam 781 

Natural  Gas,  Flow  in  Pipes,  791,  797 

Natural  Slope,  of  Earth 573 

Non-planar  Pistons 526 

Notch, Rectangular, Efflux  from, 

683,  741 
Obelisk-shaped  Vessel;  Time  of 

Emptying 744 

Oblique  Impact  of  Jet  on  Plate,    810 

Open  Channels,  Flow  in 749-797 

Orifices  in  Thin  Plate 658,  773 

Otto  Gas-engine 641,  643 


Xll 


INDEX  TO   HYDRAULICS. 


PAGE 

Overfall,    Emptying    Reservoir 

through 741 

Overfall  Weirs. . .  .677,  683,  688,  75G 
Overfall    Weirs,     Actual    Dis- 
charge   683 

Paddle-wheel  of  a  Steamer 817 

Paraboloid  as  Free  Surface 544 

Paraboloidal  Vessel 742 

Parallelopipedical       Reservoir 

Walls 555 

Pelton  Wheel  or  Hurdy-gurdy,  809 

Pendulum,  Hydrometric 753 

Perfect  Fluid,  Definition. 515 

Permanent  Gases 605,  608 

Petroleum-engine 641 

Petroleum  Pumping 708 

Piezometer 649,  657,  700 

Pipes,  Clean 708 

Pipes,  Foul 708 

Pipes,       Thickness      of,       for 

Strength 538 

Pipes,  Tuberculated 708 

Piston  Pressures 523 

Pistons,  Moving .  524 

Pistons,  Non-planar 526 

Pilot's  Tube 751,  804,  827 

Plate  between  Two  Levels  of 

Water 568,569 

Plates,  Impulse  of  Jets  on.. 801, 

805,  810 

Plates  Moving  in  a  Fluid 813 

Plates,  Resistance  in  Sea-water.  814 

Pneumatics,  Definition 518 

Poisson's  Law 621 

Poncelet's    Experiments    with  830 

Overfalls 677 

Power     Required    to      Propel 

Ships 830 

Pressure,  Centre  of 546 

Pressure  on  Bottom  of  a  Vessel,  545 
Pressure  on  Curved  Surfaces. . .  569 

Pressure  on  Sluice-gate 551 

Pressure  per  Unit-area 519 

Pressure-energy 717 

Pressure-head 650 

Principle  of  Momentum 812 

Pyramidal  Vessel,  Emptying  of,  743 
Quaker  Bridge  Dam,  Proposed,  564 

Radian,  Definition 544 

Reaction  of  a  Water- jet 798 

Rectangular  Orifices 672,  676 

Refrigerator  of  Hot-air  Engine,  640 
Regenerator  of  Hot-air  Engine,  640 
Relative  Equilibrium  of  a 

Liquid 540 

Reservoir  of  Irregular  Shape, 

Emptying  of 746 

Reservoir  Walls 554-567 

.Resistance,  Eddy-making.  .814,  828 


PA.GE 

Resistance,  Wave-making 828 

Resistance  of  Fluid  to  Moving 

Bodies 820 

Resistance  of  Fluid  to  Moving 

Plates.. 814 

Resistance  of  Ships 828 

Resistance    of    Still  Water    to 

Moving  Solids 820 

Retaining  Walls 572-580 

Righting  Couple,   of  Floating 

Body." *  597 

Ritchie-Haskell  Direction  Cur- 
rent-meter  750 

Rivers,  Flow  in 749-797 

Robinson,    Prof.,   Experiments 

on  Flow  of  Natural  Gas 797 

Robinson's  Anemometer. ..  .824,  826 

Rounded  Orifice 663 

Safety-valves 534 

Sail-boat,  Mechanics  of  the 819 

Sail-boatjMoving  Faster  than  the 

Wind 820 

Saturated  Steam,  Heaviness  .. .  628 

Saturated  Vapor 607 

Scouring  Action  of  a  Current. .  830 
Screw  Propeller,  Augment  of. .  829 

Sewers,  Flow  in 761,  765 

Ships,  Resistance  of 828 

Ships,  Stability  of 597,  599 

Short  Cylindrical  Pipes,  Efflux 

through 689,  704 

Short  Pipes 722,  723 

Short  Pipe,  Minimum  Head  for 

Full  Discharge 724 

Simpson's  Rule 603,  747,  748 

Siphons 735 

Skin-friction 695,  828 

"Slip" 829 

Slope,  in  Open  Channels 749 

Smith,  Mr.  Hamilton;  Hydrau- 
lics   689 

Smithsonian     Scale  of    Wind- 
pressures 819 

Solid  of  Revolution,  Impulse  of 

Jet  on 803 

Specific  Gravity 589 

Spren gel's  Air-pump 656 

St.    Gothard    Tunnel,    Experi- 
ments in 787 

Stability  of  Rectangular  Wall. .  554 

Stability  of  Ships 597 

State  of  Permanency  of  Flow. . .  647 

Steady  Flow,  Definition 647 

Steady     Flow,     Experimental 

Phenomena 646 

Steady  Flow  of  a  Gas 773 

Steam,  Expanding 624 

Steam.  Flow  of 781 

Steam,  Saturated,  Heaviness  of. .  628 


INDEX  TO   HYDRAULICS. 


X11I 


Steam-engine,  Examples 627 

Steam-gauge,  Bourdon 532 

Stirling's  Hot-air  Engine 639 

Stream-line 658 

Sudbury  Conduit,  Experiments, 

758,  762 
Sudden  Diminution  of  Section 

in  a  Pipe 727 

Sudden  Enlargement  of  Section 

in  a  Pipe 721 

Surface  Floats 750 

Survey,    Hydraulic,   of  Missis- 
sippi River • 754,  759 

Tachometer 750 

Temperature,  Absolute 606 

Temperature,      Influence       on 

Flow  of  Water 703 

Tension  of  Gas 519 

Tension  of  Illuminating  Gas. . .   533 

Thermodynamics 606 

Thermometers 604 

Thickness  of  Pipes 538 

Thickness  of  Pipe  for  Natural 

Gas 796 

Thin  Hollow  Cylinders 535 

Thin  Plate,  Orifices  in 658 

Throttle-valves,   Loss  of  Head 

Due  to 730 

Time  of  Emptying  Vessels  of 

Various  Forms 737-746 

Transmission    of     Compressed 

Air 786,  790 

Transporting  Power  of  a  Cur- 
rent   830 

Trapezoidal  Section  for  Open 

Channel 765 

Trapezoidal  Wall,  Stability  of..  559 

Triangular  Orifices 675 

Triangular  Wall,  ^  Stability 561 

Uniform  Motion  in  Open  Chan- 
nel   756 

Uniform  Rotation  of  Liquid 542 

Uniform  Translation  of  Liquid..  540 

Upsetting  Couple. . .  599 

Vacuum- chamber,  in  Siphon. . .  736 
Valve-gates,  Loss  of  Head  due 

to 730 

Vanes,  Impulse  of  Jets  on.  .801,  805 


I  PAGE 

'  Vapors 516,  607 

Variable  Diameter,  Long  Pipe  of,  794 
Variable  Motion  in  Open  Chan- 
nels   768 

Velocities  in  Section  of  River. .  754 
Velocity  of  Efflux  as  related  to 

Density 668 

Velocity-head 649 

Velocity  Measurements  in  Open 

Channel 750 

Vena  Contracta 659 

Venturi's  Tube 693 

Venturi  Tube,  New  Forms 694 

Venturi  Water-meter 725 

Volume  of  Reservoir  Found  by 
Observing  Time  of  Emptying,  748, 

Water,  Compressibility  of 519* 

Water-formula    for    Flow    of 

,    Gases 774 

!Water,  Heaviness,  at  Different 

i        Temperatures 518 

Water  in  Motion 646 

Water-meter,  Venturi 725 

Water-ram 530,  538 

Wave-making  Resistance 828 

Webb,  Prof.,  Experiments  on 

the  Reaction  of  Jets 800 

Wedge  of  Maximum  Thrust . . .  573 
Wedge-shaped  Vessel,  Time  of 

Emptying 742 

Weirs,  Overfall. .  .677,  683,  688, 

756,  772- 
Weisbach's  Experiments. . .  .682, 

685,  686,  691,  707,  721,  804 
Weisbach's    Experiments  with 

Elbows  and  Bends 727,  728 

Weser,  River,  Backwater  in 772 

Wetted  Perimeter 697,  749 

Wex,  von,  Hydrodynamik 688 

Whirling  Machine 816,  826 

Wind-pressure  818 

Wind -pressure,        Smithsonian 

Scale  of 819 

Woltmann's  Mill 750 

Work    of    Compressed-air  En- 
gine   631 

Work  of  Expanding  Steam 624 

Work  of  Jet  on  a  Vane 807 


DATE 


YC  I 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


